Abstract Nonsense

Crushing one theorem at a time

Actions by p-Groups (Pt. II)


Point of Post: This is a continuation of this post.

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Probably the most significant use of actions by p-groups is given by a clever proof of Cauchy’s Theorem given by McKay. Namely:

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Theorem (Cauchy): Let G be any group (not necessarily a p-group) and p a prime with p\mid |G|. Then, there exists x\in G with |x|=p.

Proof: Let X=\left\{(g_1,\cdots,g_p)\in G^p:g_1\cdots g_p=e\right\}. Since we can clearly choose the first p-1 coordinates of a tuple in X and the last is determined as the inverse of the product of the first p-1 we have that \#(X)=|G|^{p-1}. It can be easily verified that H=\mathbb{Z}_p acts on X via cyclic shifting (forward) of the coordinates of a tuple in X. More explicitly,

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k\cdot(g_1,\cdots,g_p)=\left(g_{1+k\text{ mod }p},\cdots,g_{p+k\text{ mod }p}\right)

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Indeed, it clearly suffices to prove that k\cdot (g_1,\cdots,g_p)\in X for all (g_1,\cdots,g_p)\in X since the other properties of a group action are apparent. But, this is clear by conjugation, namely

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\begin{aligned}e &=g_{p-k+1}\cdots g_p e(g_{p+k-1}\cdots g_p)^{-1}\\ &=g_{p-k+1}\cdots g_p g_1\cdots g_p(g_{p+k-1}\cdots g_p)^{-1}\\ &=g_{p+k-1}\cdots g_p g_1\cdots g_k\end{aligned}

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but, the very last expression is the product left-to-right product of k\cdot (g_1,\cdots,g_p) and so k\cdot (g_1,\cdots,g_p)\in X as desired. Note then that by the fundamental theorem we have that \#(X^H) is divisible by p (since p\mid |G|^{p-1} since p\mid |G|). But, since (e,\cdots,e)\in X we must have that \#(X^H)\geqslant p and so there exists (g_1,\cdots,g_p)\in X with (g_1,\cdots,g_p)\ne (1,\cdots,1). That said, note that since (g_1,\cdots,g_p) is fixed by every element of H we have that (g_1,\cdots,g_p)=(g_p,\cdots,g_1)=\cdots=(g_2,\cdots,g_1) and so, in particular, g_1=\cdots=g_p=g and by construction g\ne 1. But we see then that since (g,\cdots,g)\in X we have that g^p=1. It follows then that |g|\mid p and since |g|\ne 1 we have that |g|=p as desired. \blacksquare

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From where we get the following, fairly relevant corollary:

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Theorem: A finite group G is a p-group if and only if |g| is a power of p for every g\in G.

Proof: Evidently from Lagrange’s theorem we have that if G is a p-group then every element of G has prime power order. Conversely, evidently if p,q are two different primes dividing |G| then by Cauchy’s Theorem we know that there exists x,y\in G with |x|=p and |x|=q. It follows then that every element of G has order equal to the power of a single prime. \blacksquare

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And  a simple proof that the only finite fields can be of order p^n for some prime p. Indeed:
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Theorem: Let F be a finite field, then |F|=p^n for some prime p.

Proof:  Note that the map f_{a,b}:F\to F:x\mapsto ab^{-1} is an (additive) isomorphism for each a,b\in F^\times and so |a|=|b| for all a,b\in F^\times. Note though that if |F| is divisible by two primes, we could find two elements of different orders. Thus, |F| is a primer power as desired. \blacksquare

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What we now wish to do is prove that \left[G:N_G(H)\right]\equiv \left[G:H\right]\text{ mod }p if H is a p-group sitting inside the larger group G. From where we will be able to prove that if H\notin\text{Syl}_p(G) then H\ne N_G(H).

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Theorem: Let G be any finite group and H a p-subgroup. Then \left[N_G(H):H\right]\equiv \left[G:H\right]\text{ mod }p.

Proof: Consider the action of H on X=G/H given by left multiplication.  What we’d like to figure out is what X^H really is. Well, note that if gH\in X^H then hgH=gH for all h\in H and so hg=gh' for some h'\in H and so g^{-1}hg\in H. Thus, since h was arbitrary we may conclude that g^{-1}Hg=H and so g\in N_G(H) and conversely one can check that if g\in N_G(H) then gH\in X^H. Thus, X^H=\left\{gH:g\in N_G(H)\right)=N_G(H)/H. Thus, \#(X^H)=\left[N_G(H):H\right]. The conclusion then follows from the fundamental theorem. \blacksquare

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Corollary: If p\mid [G:H] then p\mid [N_G(H):H], and since [N_G(H):H]\ne 0 we may conclude that \left[N_G(H):H\right]\geqslant p and so N_G(H)\supsetneq H.

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This tells us that if |G|=p^n and H\leqslant G has |H|=p^{n-1} then H\unlhd G, since by the corollary we have that H\subsetneq N_G(H) and so clearly N_G(H)=G. That said, we already knew this from the theorem that says if the index of a subgroup is the smallest prime dividing the group then it’s normal. What this theorem is really useful for is:

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Theorem: Let G be a finite p-group with |G|=p^n. Then, there exists a chain of subgroups 

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\{e\}=H_0\unlhd\cdots\unlhd H_{n-1}\unlhd H_n=G

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with |H_j|=p^j.

Proof: We first use Cauchy’s Theorem to prove that there exists some x\in G with |x|=p, we then take H_1=\langle x\rangle. We then proceed by ‘induction’. Suppose that we have found a group H_j with j<n and |H_j|=p^j. We see then that [G:H_j]=p^{n-j}\geqslant p and so by the previous corollary we have that H_j\subsetneq N_G(H_j). We see then that p\mid |N_G(H_j)/H| and so by Cauchy’s theorem we have that there exists gH_j\in N_G(H_j) with |gH_j|=p. But, from the correspondence theorem we know that \langle gH_j\rangle=K/H_j for some K\leqslant N_G(H_j) and moreover that \left[K:H\right]=\left|\langle gH_j\rangle\right|=p and so |K|=p^{j+1}. To see that H_j\unlhd K we may merely appeal to the paragraph immediately preceding this theorem. Thus, the ‘induction’ is complete. \blacksquare

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This allows us to conclude from the first (weak) Sylow’s Theorem, namely the existence of Sylow p-subgroups, the stronger result that if p^m\mid |G| where G is any group that G contains a subgroup H with |H|=p^m.

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To finish this post let’s give a cool proof of Wilson’s theorem:

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Theorem(Wilson): Let n\in\mathbb{N}. Then, (n-1)!\equiv -1\text{ mod }n if and only if n is prime.

Proof: Suppose that n=p is prime. Consider the action of \mathbb{Z}_p=G on X=\left\{(g_1,\cdots,g_p)\in S_p^p:g_1\cdots g_p=e\right\} as previously discussed. We noted that X^G is equipotent to \left\{g\in S_p:g^p=e\right\} and so by the Fundamental Theorem we have that \{g\in S_p:g^p=e\}\equiv 0\text{ mod }p. Note that if g\in S_p is non-identity then it’s order is the least common multiple of the cycle lengths of its disjoint cycle decomposition. It clearly follows that the only non-identity elements of S_p of order p are the p-cycles. Now, how may distinct p-cycles are there? Well, there are p! ways to arrange 1,\cdots,p,  but each of these arrangements is equivalent to precisely p others (the p cyclic shifts forward) and so the number of distinct p-cycles is (p-1)!. Thus, since the only g\in S_p for which g^p=e are the elements of order p and e we may conclude that \#\{g\in S_p:g^p=e\}=(p-1)!+1, thus as previously noted (p-1)!+1\equiv 0\text{ mod }p.

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Conversely, if n has a proper divisor d then we have to have that d\mid (n-1)! but by assumption it would also divide (n-1)!+1 and so d\mid 1, which is a contradiction. \blacksquare

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September 15, 2011 - Posted by | Algebra, Group Theory | , , , , , , , , , ,

1 Comment »

  1. […] group. Suppose then that we have found a subgroup of with for . We know then that and so by a previous theorem we have that . Thus, by Cauchy’s Theorem there exists with . By the fourth isomorphism […]

    Pingback by Sylow’s Theorems Revisited « Abstract Nonsense | September 20, 2011 | Reply


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