Explanation: For any action, you can define a permutation. It is easy to show this action is injective. To show that it's surjective note that for any homomorphism, you can build an action based on it.
We will call a permutation representation any homomorphism from a group into any symmetric group.
For a group G acting on set A we define the orbit of an element \(a\in A\) to be the set of all \(b\in A\) such that \(a\sim b\) where this means that there exists some \(g\in G\) such that \(a = g\cdot b\).
Explanation: It's very straight-forward to show the relation is an equivalence relation. So the interesting task is to show that the orbit of an element a has size \(|G:G_a|\). The intuition is the same as for Lagrange's Theorem. In that theorem you imagine a subgroup H, and then imagine multiplying it by any element. You get a coset xH and only one of the following two is possible: H = xH or \(H\cap xH = \emptyset\). So all these cosets partition and have the same size, so the size of the subgroup is |G| over the number of cosets.
In this case we have orbits and the stabilizer group. We want to identify each coset of \(G_a\) with some element of the orbit \(\mathcal O_a\). So we set up the bijection \(\varphi:\mathcal O_a \rightarrow \{xG_a:x\in G\}\) defined on any \(x\in\mathcal O_a\) in the following way. Let \(g\in G\) such that \(g\cdot x = a\). Then \(\varphi(x) = gG_a\).
This is injective since, for any two \(x,y\in G_a\) such that \(g_1\cdot x = g_2\cdot y = a\) then \(\varphi(x)=\varphi(y) = g_1G_a = g_2G_a\) entials that \(g_2^{-1}g_1 \in G_a\). So \( g_2^{-1}g_1\cdot a = a \) and so \(g_1\cdot a = g_2\cdot a = x = y\). Why is the map surjective? If \(gG_a\) is any coset, we need to find an \(x\in \mathcal O_a\) such that \(g\cdot x = a\). This is the same as \(x=g^{-1}\cdot a\) and so we've found x.
Explanation: Take any \(\sigma \in S_n\) and consider the subgroup it generates, call it \(G = \langle \sigma \rangle\). Also consider how G acts on A = {1, 2, ..., n}. From the above, A is partitioned into orbits of the elements of G. Those orbits make up the cycles, which we know since G is cyclic and it's not hard to prove that inside each cycle we can get from any element to another with a sufficient power of \(\sigma\). This shows that \(\sigma\) has a cycle decomposition.
To see that it's unique, note that in any orbit the order of the elements is determined by the powers of \(\sigma\) up to the choice of the first element (i.e. the orbit representative). Any other rearrangement comes from the order in which the orbits are considered. But the orbits themselves, and the sequence in which elements are visited within each, is unique and hence so are the cycles and the order in which their elements appear.
We can view any given element of a group as acting on the group by multiplication. We will later see that this action is both transitive and faithful, and the stabilizer of any point is always just the identity.
In a way it is even more interesting to act on cosets. Note that when the subgroup is just the identity then the cosets are just the individual elements of the group. So in this case we might as well act on the group (as we described it in the previous paragraph).
If \(H \leq G\) then we homomorphically map G into a symmetric group. In particular if there are n cosets of H (i.e. if |G:H| = n) then we map into \(S_n\). The map is defined by \(g \mapsto \sigma_g\) where \(\sigma_g\) sends i to j if g sends the coset indexed by i to the coset indexed by j. If this homomorphism from G to \(S_n\) is called \(\pi_H\) then we call this the permutation representation associated with H. (The system of indexing is unimportant.)
Explanation: To show the action is transitive we need to see that you can "get from any coset to another by multiplication". Of course to get from aH to bH you can multiply by \(ba^{-1}\).
The stabilizer of H is the set of all elements that fix H. The previous chapter makes it pretty clear that everything in H fixes H and everything not in H doesn't.
The kernel of the action is the set of all group elements that fix every G-set element. That is to say, we need all elements that fix all cosets. $$ gxH = xH $$ $$ x^{-1}gxH = H$$ implies \(x^{-1}gx \in H\) and so \(g\in xHx^{-1}\). We can turn this observation into a bi-directional proof to show that the sets are equal.
Finally, let \(\pi_H\) be the permutation representation associated with H. Before anything else we should actually confirm that this is a homomorphism. If \(a,b\in G\) then \(\varphi(ab) = \sigma_{ab}\). The equality that we want to proof is an equality of the functions \(\sigma_{ab}=\sigma_a\sigma_b\) so we need to show that these map the same things to the same things.
Let iH be the coset indexed i, and \(\sigma_b(i) = j\). That means \(b\cdot iH = jH\). Likewise let kH be the coset such that \(\sigma_a(j)=k\) so that \(a\cdot jH = kH\). $$ \sigma_{ab}(iH) = ab\cdot iH = a\cdot jH = kH $$ $$ (\sigma_a\sigma_b)(iH) = \sigma_a(jH) = kH $$
Moving on, we want to show that ker(\(\pi_H\)) is the largest normal subgroup of G such that \(ker(\pi_H)\leq H\). We already know that the kernels of homomorphisms are normal. Checking that ker(\(\pi_H\)) is in H is straight-forward. This is the set of elements which map to the identity permutation. That makes it the set of elements which fix all cosets (and so is the same as the set we discussed above). Anything which fixes H in particular is in H.
So we want to show that any normal subgroup which is contained in H is contained in ker(\(\pi_H\)). From the previous observation that \(\displaystyle ker(\pi_H) = \bigcap_{x\in G}xHx^{-1}\), together with the fact that normal subgroups are fixed under conjugation, this is immediate.
Explanation: Apply the above to the subgroup of just the identity. Since the kernel is the identity the map is injective.
Explanation: For the case when the prime is 2, the proof goes by directly thinking about the cosets. One coset is the subgroup itself. If you think of left and right cosets, they both share the subgroup as a coset. Therefore everything that remains is also "the other coset". Therefore all left cosets are the same right coset, and therefore the subgroup is normal.
For a more general group of order |G:H| = p the least prime dividing |G|, we know that this is homomorphic to a subgroup of \(S_p\). In particular we'll try to see that ker(\(\pi_H)\) is just the identity, and we'll do this by scrutinizing its size. Then we'll have that the homomorphism is an isomorphism with its image.
Now from a couple theorems ago that \(\ker \pi_H \leq H\leq G\). And we know that all of these groups are finite, so |G:H|=|G|/|H| and \(|H:\ker \pi_H| = |H|/|\ker \pi_H|\) and \(|G:\ker \pi_H|=|G|/|\ker \pi_H|\). So $$ |G:K| = |G:H||H:K| $$ $$ |G:K| = pk $$ where k = |H:K|.
Now since we're mapping G into \(S_p\) and since \(\ker \pi_H\trianglelefteq G\), then by the First Isomorphism Theorem \(G/\ker \pi_H\) is isomorphic to the image, which is a subgroup of \(S_p\). So \(pk=|G/K|\) divides p! and so k|(p-1)!. That shows that k must be made of factors that do not exceed p-1.
Now notice that k | |G| so if any prime divides k then it divides |G|. So if we have any prime dividing k it must be greater than or equal to p (by the minimality of the choice of p). So k = 1.
The orbits of the action of a group on itself by conjugation are called conjugacy classes. Two elements in the same orbit are called conjugates in G. Just like last time we can generalize this notion to acting on cosets of a subgroup.
Explanation: Note that, by a previous theorem, the size of an orbit is the index of the stabilizer. But also the stabilizer when acting by conjugation is the normalizer. Hence the number of conjugates of a subset is equal to the index of the normalizer.
The claim for a single element is a consequence of the above when we take the set to be a singleton.
Explanation: The class equation is just a sum over all conjugacy classes. The ones that are singletons are just elements in the center. Everything else has the size that we discussed earlier, the index of the centralizer.
Explanation: In the class equation, the order of the group is on the left and is divisible by the prime. Therefore the quantity on the right is. But every orbit has index which divides the order of the group, so each term here is divisible by the prime. Hence the order of the center must be divisible by the prime, and so is non-trivial.
Explanation: By the previous theorem the center is not trivial. Then P/Z(P) has order 1 or p but in either case it's cyclic. Therefore P is abelian.
Now for the more particular part. If it has an element of order \(p^2\) we're done. Otherwise every element has order p or 1. Take one non-identity element, and another not in the set that the first one generates. These two generate a group larger than the one that they generate individually, so they together generate the whole group. It is now possible to set up an explicit isomorphism from \(\langle x\rangle \times \langle y\rangle \) to \(Z_p\times Z_p\).
Conjugation for matrices is basically the "change of basis" from matrix to another. For permutations the situation is analogous.
Explanation: If \(\sigma=(a_1 \ a_2 \ \dots \ a_p)(b_1 \ \dots \ b_q)\dots \) then the theorem claims that \(\tau\sigma\tau^{-1}\) is $$(\tau(a_1) \ \tau(a_2) \ \dots \ \tau(a_p)) (\tau(b_1) \ \dots \ \tau(b_q)) \dots $$
The proof goes entirely by saying "Well if the theorem is true, and if \(\sigma\) sends some i to j, then let's see what \(\tau\sigma\tau^{-1}\) does to input \(\tau(i)\)."
We've already seen that conjugates have the same cycle type. Now suppose \(\sigma, \tau\) have the same cycle type. Then they have the same number of cycles each with the same length. Pairing these up, we can construct a permutation by the pairing of the elements in these matching cycles. Conjugating \(\sigma\) by this permutation produces \(\tau\) which can be seen easily by the previous theorem.
From having proved this first part we now know that the conjugacy classes of \(S_n\) just is the partition of the group by cycle types. Every cycle type corresponds to a partition of n, so we have the theorem.
... Wait what? How is there a cycle type per permutation of n? This requires some more explanation. Make the alignment like this: For any given cycle type, the length of each factor cycle corresponds to a term in the partition.
Explanation: We could have started this proof a long time ago, but we would have stalled out at a certain point. But with the help of the theorem above, we can complete the proof. We start by observing that of course any m-cycle commutes with all of its powers, and there are m of these. Moreover, of course, it commutes with anything that is disjoint from it, and there are (n-m)! of these. And of course, it commutes with any product of the two, and there are m(n-m)! of these.
The question where we would have stalled out is: Are there any other elements which commute with the given m-cycle? We can answer this question using what we now know about its centralizer and the relationship to its conjugacy class. We know that its conjugacy class is the set of all permutations of the same cycle type. And we know that, if we call the permutation \(\sigma\), the conjugacy class has size equal to \(|S_n:C_{S_n}(\sigma) = \frac{n!}{|C_{S_n}(\sigma)}\).
Being an m-cycle it has a particularly easy to cycle type. There are \(\frac{n(n-1)\cdots(n-m+1)}{m}\) many m-cycles (the total number of ways to pick coordinates without replacement, divided by the number of "rotations" of these). Putting all of the information we've gathered above together, $$ |C_{S_n}(\sigma)|=m(n-m)! $$
What this tells us is that there are no other conjugates of \(\sigma\) than the ones we counted at the start! Note that another way of understanding what this is saying, is that $$ C_{S_n}(\sigma) = \{\sigma^i\tau| 0\leq i < m, \text{ and } \tau \in S_{n-m}\} $$
Explanation: To show that \(A_5\) has no proper non-trivial normal subgroup, we will make use of the following general observation: For a normal subgroup, an element inside of it is never sent outside by conjugation; and an element outside is never sent inside. Put more formally, if H is normal then \(x\in H\) if and only if \(gxg^{-1}\in H\) for every \(g\in G\). Part of this we already know immediately because we know that normal subgroups are closed under conjugation. In the other direction, if we suppose that some conjugate of x is outside H then x must be outside of H. For x were inside H this would entail that it's "inverse conjugate" sent it outside, which we know is impossible.
Note that another way of understanding what we've just learned is that every normal subgroup is the union of the conjugacy classes inside of it. Because each conjugacy class is either entirely inside or entirely outside.
Since this is a property of all normal subgroups, we'll show that there cannot be any subgroup with this property. And to do that, we'll work out the sizes of all the conjugacy classes in \(A_5\). As a note about how we do this, keep in mind that two elments in \(A_5\) may be of the same cycle type but may not be conjugate to each other in \(A_5\), because \(A_5\) is not all of \(S_5\). The element that conjugates one permutation to another of the same cycle type may be missing.
Now notice that every cycle type in A5 is represented by the elements (1), (1 2 3), (1 2 3 4 5), and (1 2)(3 4). And although being of the same cycle type doesn't entail being conjugate, being conjugate certainly still entails being of the same cycle type. So every conjugacy class is made up of the same cycle type--there just might be multiple conjugacy classes of the same cycle type.
Let's start by analyzing the 3-cycles. We know the only things in the centralizer of (1 2 3) are powers of this, times the elements which are disjoint. But the only disjoint element in A5 is (1) so the centralizer is just <(1 2 3)> and this has size 3, hence its conjugacy class has size 60/3 = 20. With the same sort of combinatorial analysis as we've been using, you can see that this is exactly the same as the total number of 3-cycles, so we've found that all 3-cycles are conjugate.
Now let's analyze the 5-cycles. Again the only thing disjoint is (1) so the centralizer is <(1 2 3 4 5)> which has order 5 and therefore conjugacy class of size 60/5 = 12. However, there are 24 different 5-cycles. So what's in the conjugacy class of (1 2 3 4 5) and what's not? An example of something that's not is (1 2)(1 2 3 4 5)(1 2) = (2 1 3 4 5). Clearly that's because (1 2) is not in A5. In general a 5-cycle is not in the conjugacy class of (1 2 3 4 5) if it is conjugate by an odd permutation. Since half of the 5-cycles are conjugate by even permutations and half by odd, then this is how the conjugacy classes split up. There are two conjugacy classes of 5-cycles and there are 12 elements in each.
Finally there are 60-20-24-1 = 15 different permutations in A5 which have type (1 2)(3 4). For this, a brute-force check shows that (1 2)(3 4) commutes with every element of the same cycle type and nothing else, so that |CA4((1 2)(3 4))| = 4 and therefore the conjugacy class contains 60/4 = 15 which means it contains all 15 elements of the shared cycle type.
Ok, phew! Nowe we know all the conjugacy classes and they have sizes 1, 12, 12, 15, and 20. If any group were normal, from what we discussed at the start, it would have to have size a sum of these numbers. But since non sum of these numbers divides 60, this is impossible.
An automorphism is an isomorphism from a group to itself. The set of all automorphisms of a group G is denoted Aut(G).
Explanation: The proof is routine.
Moreover, an automorphism is an element of the group of permutations on G viewed as merely a set.
Explanation: It is very straight-forward to verify the first claim. We can see that through conjugation by g we get a group action on H, and this action affords a permutation representation \(\pi(g)=\varphi_g\in S_H\). The kernel of \(\pi\) being the set of group elements which map to the identity map, that means these group elements fix every \(h\in H\) under conjugation, which is equivalent to commuting.
Note that the above proposition shows us that normal subgroups are structurally the same as their conjugates.
Explanation: For any group G we know that \(G\trianglelefteq G\). So conjugation by any g is an isomorphism and therefore its restriction to any subgroup K is also an isomorphism for K.
Explanation: \(H\trianglelefteq N_G(H)\) for the first claim, and \(G\trianglelefteq G\) for the second.
Conjugation by g is called an inner automorphism of G. The set of all inner automorphism is a subgroup of the group of automorphisms, and is denoted Inn(G). A subgroup is characteristic, denoted H char G, if it is fixed by every automorphism of G. Characteristic groups are always normal. If a subgroup is the unique group of a given order then it is characteristic. And K char H together with \(H\trianglelefteq G\) entaisl \(K\trianglelefteq G\). In this way they can be thought of as "strongly normal".
Explanation: We build an isomorphism \(\Psi: Aut(Z_n)\to (Z_n)^\times\) by taking any automorphism \(\psi\) and chase the generator. That is to say, \(\psi(1)=a\), and for no other automorphism can this be true. Moreover since \(\psi\) is an automorphism then 1 and a have the same order, so (a,n) = 1. In fact, this is an if and only if: For every a coprime to n there is an automorphsim mapping 1 to a. So $$\Psi(\psi) = \psi(1) \mod n $$
Let p be a prime, and \(p^\alpha\not\mid m\). We define:
Explanation: For the first claim, we proceed by induction on the order of G. When |G| = 1 there is nothing to prove. Now with |G| = pαm > 1, we divide into two cases: Either p | Z(G) or not. If it does then by Cauchy's theorem for abelian groups, Z(G) has a subgroup N of order p. Apply the inductive hypothesis to G/N. We can use this with the isomorphism theorems to wrangle the subgroup we want.
Next we handle the case when p ∤ |Z(G)|. From the class equation there must exist at least one CG(gi) such that p does not divide its index. But if it doesn't divide the index it must divide m instead (prime facts). So now CG(gi) has a Sylow p-subgroup by the induction clause, P. Of course |P| = pα so it is a Sylow p-subgroup of G.
Alright moving on, we build this in many stages. Let Q be any p-subgroup of G. List out all the conjugates of P and call it \(\mathcal S\). So \(\mathcal S = \{P_1, \dots, P_r\}\) with each Pi = giPgi-1. Suppose for contradiction that Q is not contained in any of them, so that $$ |Q : Q\cap P_i| > 1 $$
Notice at this point that this is enough to infer, from the orders of the groups involved, that p | |Q : Q∩Pi| for every i.
Now at this point we need to go on an extended lemma to show that \(|Q:Q\cap P_i|\) is the size of the orbit of P i under conjugation by Q. So to get started, first let \(\mathcal O_1, \mathcal O_2, \dots, \mathcal O_s\) be the partition of \(\mathcal S\) into the orbits of \(\mathcal S\) under conjugation by Q. That is to say each \(\mathcal O_i\) is the set of all qPjq-1 for some Pj, and also \(\mathcal S = \bigcup \mathcal O_i\), and also \(r = \sum |\mathcal O_i|\). Moreover, re-number the Pi such that the first s of them are representatives of their orbits. Then for each 1 ≤ i ≤ s we have \(P_i\in \mathcal O_i\).
Now with all of this notation in place we can state more formally what we're trying to show. We're trying to show \(|\mathcal O_i| = |Q:Q\cap P_i|\). From the start of the chapter we know that the size of an orbit is the index of the stabilizer. Hence \(|\mathcal O_i| = |Q:Q_{P_i}|\). Because the action is conjugation by elements in Q that makes QPi the same as NQ(Pi). It is trivial to show that NQ(Pi) = NG(Pi)∩Q.
Now in fact we have to get even further diverted from our main proof. We want to show that NG(Pi)∩Q = Pi∩Q. This will in fact be a lemma.
Explanation: We already know that any group is contained in its normalizer, so half of the equality is trivial. What takes work is \(Q\cap N_G(P)\subseteq Q\cap P\), which will follow if we can show that \(Q\cap N_G(P) \leq P\). The shape of the proof will be to show that P(Q∩NG(P)) is a p-subgroup of G. Since P itself is a p-subgroup of largest possible order, and is contained in P(Q∩NG(P)), then P(Q∩NG(P)) = P which entails Q∩NG(P) ≤ P.
So now the lemma hinges on showing P(Q∩NG(P)) is a p-subgroup. Since Q∩NG(P) is in the normalizer of P, we have a theorem saying that the product is always a subgroup. Analyze the order of the product.
Ok! Finally we can get back to proving Sylow's theorem, now that we know \(|\mathcal O_i| = |Q:Q\cap P_i|\). From what we established earlier we now have the two facts that \(p\mid |\mathcal O_i|\) and \(\sum |\mathcal O_i|=r\) hence we infer that p | r. This is the fact that we will contradict, in particular by showing that r ≡ 1 (mod p).
To show this congruence relation, we re-do several things above that we did with Q, except that we do it with P in place of Q. So this time call \(\mathcal O_1', \mathcal O_2', \dots, \mathcal O_t'\) the orbits of \(\mathcal S\) under conjugation by P. If in particular we choose the labeling where P1 = P then we have \(|\mathcal O_1 | = 1\). Yet the rest of the argument goes through just as before, where every other orbit is still divisible by p. Hence r ≡ 1 (mod p) and we have our contradiction. Namely, the second claim in Sylow's theorem is now proved.
We move on to the third claim, which in fact we've already proved when we showed r ≡ 1 (mod p). For at this point we also know that r = np, since obviously the Sylow p-subgroups are the conjugates of P.
And finally since \(n_p=|\mathcal O_1| = |G:G_P| = |G:N_G(P)|\) we have the last statement.
Explanation: The equivalence (1) ⇔ (2) pretty much just comes from definitions and the fact that all Sylow p-subgroups are conjugate.
The equivalence (2) ⇔ (3) is half-done from the fact that every characteristic subgroup is normal, and the other half owes to the equivalence between (2) and (1).
The equivalence (4) ⇔ (1) is half done directly by using the fact that every p-subgroup is in a conjugate P but P is normal. In the other direction, union over all elements of order a p-power, which must contain P, but P is maximal.
Note that if p is a prime not dividing the order of a group, technically according to the definition, |G| = pαm where α = 0 and |G| = m. So the Sylow p-subgroup is the group of order pα = 1 which must be the trivial group. Also if |G| = pα then the only way to write |G| = pβm where p ∤ m is when m = 1 and so G is the Sylow p-subgroup.
Note that since all Sylow p-subgroups are conjugate, then in an abelian group, every Sylow p-subgroup is unique. Since in this case we can take the subgroup of all elements of order p and show that this is a Sylow p-subgroup, then this is THE Sylow p-subgroup.
It is also worth noting that because np is the index of the stabilizer, it divides the the order of the group. Since it cannot divide p that means it divides m.
Explanation: Suppose |G| = pq where p, q are primes and p < q. Then nq | p and p < q so nq < q. But also nq ≡ 1 (mod q), so nq = 1.
Explanation: Under this further condition we have that np = 1 and so there is a unique Sylow p-subgroup, call it P. And there is a unique Sylow q-subgroup, call it Q. Each of these has order a prime number and therefore must be cyclic (pretty sure we proved that earlier). From here we use some fairly fancy footwork to show that P is contained in the center of G.
First we refer to the fact from the discussion of automorphisms: \(N_G(H)/C_G(H)\) is congruent to a subgroup of Aut(H). Applying this to P and using the fact that P is normal, therefore its normalizer is the whole group, then \(G/C_G(P)\leq Aut(P)\) (said roughly). Because P is cyclic of order p then it might as well be Zp. Again using automorphism facts, \(Aut(Z_p)\cong (Z_p)^\times\) and so \(G/C_G(P) \leq (Z_p)^\times\) which has order p-1. So $$ |G|/|C_G(P)| = pq/|C_G(P)| \leq p-1 $$ $$ |C_G(P)| \geq \frac{pq}{p-1} > q$$ The only possible order left for \(C_G(P)\) is pq, so P commutes with everything in G. But now we can infer that the order |xy| = lcm(p,q) = pq, so that G = ⟨xy⟩.
Explanation: Note that a subgroup of order 15 has index 2, so is normal for that reason. Also notice that the previous result applies too, with p=3, q=5. In fact, according to the Sylow theorems, n3 ≡ 1 (mod 3) and n3 | 5. This could only be n3 = 1 and hence there is a unique normal subgroup of order 3. We now have the full force of the previous result, and can infer that any group of order 15 is cyclic.
All that remains then is to show the existence of any subgroup of order 15 in a group of order 30. Call the normal Sylow 3-subgroup P, and any Sylow 5-subgroup Q. Since we already know P ⊴ G then PQ is a subgroup of order 15.
Explanation: Assume there is no normal Sylow 3-subgroup and show that G ≅ A4. Like before we start with an analysis of the numbers of Sylow groups, and show n3 = 4. Now if A, B are two distinct subgroups of order 3, then any element in either of them will generate the whole group. Hence their intersection can only be the identity. So each has 2 elements not shared with any others, so this accounts for 1 + 4*2 = 9 elements among them.
Now on a different track, we can observe that |G : NG(P)| = n3 = 4 = 12/|NG(P). Since the order of NG(P) is 3 and P has order 3, and P ≤ NG(P), we have equality.
Next we examine the permutation representation afforded by G acting by conjugation on the four Sylow 3-subgroups. The kernel of this action fixes P under conjugation, and since P is not normal in G, this kernel lacks at least one element. Since the kernel is also contained in the normalizer, which we showed was equal to P, the only possible subgroup is the trivial subgroup. That is to say, the permutation representation is an injective map φ: G → S4.
This makes G isomorphic to a subgroup of S4. Moreover, A4 has all 8 of the elements of order 3 in S4, so G maps into a subgroup of A4. Moreover, this subgroup has at least 9 elements, and |A4|=12, so in fact it is all of A4.
Explanation: If p < q then quick analysis of the number of Sylow subgroups yields P ⊴ G. Otherwise we can show that if Q is not normal in G then by the divisibility criteria nq must be p or p2. From the congruence criteria and the assumption that Q is not normal, we can infer that q < nq and from p < q we know nq cannot be p. So q = p2.
Now write out the congruence relation by definition, use the above, rearrange, and infer that q|p-1 or q|p+1. Of course the former is impossible from p < q, but the latter entails q=p+1. The only adjacent primes are 2 and 3. Now apply the previous result.
Explanation: By way of contradiction, suppose both n5 > 1 and that there is a normal subgroup H ≠ 1 or G. Then by examining the congruence and division criteria, n5 = 6. By a now familiar analysis of the size of the normalizer, it must have size 10.
The possible orders of H are 2, 3, 4, 5, 6, 10, 12, 15, 20, or 30. We eliminate all of these possibilities. We can eliminate 5, 10, 15, 20, and 30 at the same time, since if 5 | |H| then there is a Sylow 5-subgroup in H. Since H is normal it contains all 6 conjugates of this group. This implies at least 25 elements in H so already |H| = 30, but yet a subgroup of order 30 has index 2, which must be normal.
By the previous theorems, if |H| = 6 or 12 then H has a PROPER characteristic subgroup which therefore is normal in G. If we now show that normal subgroups of orders 2, 3, and 4 are impossible, we will have eliminated all possibilities.
So now suppose that H has a subgroup of order 2, 3, or 4. We will look at the quotient group, and infer from its order that the quotient group has a proper normal subgroup with order 5. From the third isomorphism theorem this entails that this subgroup of the quotient group corresponds to a subgroup of G, with order divisible by 5. Since we've eliminated that possibility, there can be no such subgroup.
Explanation: ⟨(1 2 3 4 5)⟩ and ⟨(1 2 3 5 4)⟩ are distinct Sylow 5-subgroups.
Explanation: For any simple group of order 60, start by considering the number of Sylow 2-subgroups. The only possibilities, by non-normalcy, divisibility, and congruence criteria, are n2 = 3, 5, or 15.
Next we prove that no proper subgroup has INDEX less than 5. That means indices of 2, 3, or 4, and therefore subgroups of order |H| = |G|/|G:H| = 60/2 or 60/3 or 60/4. In all cases, we have previous theorems showing that groups of this order have a proper non-trivial normal subgroup.