# Cauchy's Theorem question, abelian case

• Nov 5th 2012, 08:33 PM
amyw
Cauchy's Theorem question, abelian case
Theorem: Let G be a finite group and p be a prime. If p divides the order of G, then G has an element of order p.
Proof 1: We induct on n = |G| and consider the two cases where G is abelian or G is nonabelian. Suppose G is abelian. If G is simple, then it must be cyclic of prime order and trivially contains an element of order p. Otherwise, there exists a nontrivial, proper normal subgroup http://upload.wikimedia.org/math/5/b...fa6b461b0e.png. If p divides |H|, then H contains an element of order p by the inductive hypothesis, and thus G does as well. Otherwise, p must divide the index [G:H] by Lagrange's theorem, and we see the quotient group G/H contains an element of order p by the inductive hypothesis; that is, there exists an x in G such that (Hx)p = Hxp = H. Then there exists an element h1 in H such that h1xp = e, the identity element of G. It is easily checked that for every element a in H there exists b in H such that bp = a, so there exists h2 in H so that h2 p = h1. Thus h2x has order p, and the proof is finished for the abelian case.

Could someone please explain for me using small words and step by stem everything after the word 2nd otherwise...
Does index [G:H] mean the number of (right) cosets of H in G? Also, where it says, it is easily checked...how?

We have gone over this proof in class 4 times. Also, I have gone to office hours twice. I understand it and follow when I'm there, but I can't replicate.
• Nov 6th 2012, 08:50 AM
Deveno
Re: Cauchy's Theorem question, abelian case
ok, for a finite group G, and any subgroup H:

[G:H] = |G|/|H| (we know this is an integer from Lagrange's theorem).

if H is a normal subgroup, then G/H is a group, and since |G/H| is the number of cosets of H (these are the elements of G/H), we have:

|G/H| = [G:H] = |G|/|H|.

so, for any group G with a normal subgroup H:

|G| = |H|*|G/H| (and even if H is not normal, we STILL have |G| = |H|*[G:H]).

now if p is a prime number, with p|ab, then either p|a or p|b (this is a defining property of prime numbers).

so if p divides the order of G, it has to divide one of |H| or |G/H| (maybe both, it could happen).

now if p divides |H|, we're good, since |H| is assumed less than |G|, and we can apply our induction hypothesis.

but maybe it doesn't. in that case p HAS to divide |G/H|. since |H| < |G|, and |G| = |H|*|G/H|, it follows that |G/H| < |G| as well, so we can apply our induction hypothesis to G/H (a smaller group than G).

of course, the elements of G/H are cosets of H, so an element of G/H of order p looks like this:

xH in G/H with: (xH)p = H (H is the identity element of G/H).

but (xH)p = (xp)H, so we have xp is an element of H (since (xH)p = (xp)H = H).

of course, that doesn't appear to be much help...we want an element g of G for which gp is in the subgroup {e}, not H.

so let's look at H for a second. let's take a typical element h of H. what can we say about it?

well, we know that h DOESN'T have order p. this means that gcd(|h|,p) = 1. this means that hp generates <h> just like h does.

that is, for EVERY h in H, hp generates <h>.

now ANY element h of H lies in the subgroup <h> of H. and the union of all these cyclic subgroups of H is H itself.

that is: if a is in H, a lies in <h> for some h in H (we might have to take h = a, but that's OK).

but <h> = <hp> so a = (hp)k, for some h in H (which again, might be a, might not).

thus a = (hk)p, a is a p-the power of hk. clearly if h is in H, so is hk,

so ANY element of H is the p-th power of some element of H (we can do this because p DOESN'T divide |H|, not even "partially", because p is PRIME).

ok, so we know we have this x in G with xp in H.

so xp = h, for some h in H. thus h-1xp = e.

now h-1 is in H, so there is some element (let's call it h') with:

h'p = h-1.

here is where the abelian part comes in:

consider h'x.

(h'x)p = (h'x)(h'x)....(h'x) (p times).

= (h')(h')....(h')(x)(x)....(x) (each factor p times)....because G is ABELIAN.

= (h')pxp = h-1xp = e, so h'x has order p, as desired.

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