# Problems understanding proof of Cauchy theorem

1. Feb 23, 2015

### PsychonautQQ

Here is part of the proof via wikipedia:

"We first prove the special case that where G is abelian, and then the general case; both proofs are by induction on n = |G|, and have as starting case n = p which is trivial because any non-identity element now has order p. Suppose first that G is abelian. Take any non-identity element a, and let H be the cyclic group it generates. If p divides |H|, then a|H|/p is an element of order p. If p does not divide |H|, then it divides the order [G:H] of the quotient group G/H, which therefore contains an element of order p by the inductive hypothesis...."

I am confused why if p does not divide |H| that it must divide the index of H in G. Can anyone help me out here?

2. Feb 23, 2015

### BruceW

I think it is because they have already made the assumption that p divides |G| and therefore (since p is prime), it must divide either |G|/|H| or |H|.

3. Feb 23, 2015

### PsychonautQQ

Okay, so p is a prime that divides |G|, so it must divide either |G|/|H| or |H|? Why?

4. Feb 23, 2015

### jbunniii

This is a general fact: if $a$ and $b$ are integers and $p$ is a prime number which divides $ab$, then $p$ divides either $a$ or $b$.

Proof: suppose $p$ divides $ab$ and $p$ does not divide $a$. This means that $p$ and $a$ are relatively prime. Therefore, there exist integers $m$ and $n$ such that $np + ma = 1$. Therefore, $npb + mab = b$. Since $p$ divides $ab$, it divides the left hand side of the equation, and therefore it also divides the right hand side, i..e. $p$ divides $b$.

Alternative proof: $a$ and $b$ have unique prime factorizations, say $a = p_1 \cdots p_j$ and $b = q_1 \cdots q_k$. Then the unique prime factorization of $ab$ is $(p_1 \cdots p_j)(q_1 \cdots q_k)$. If $p$ does not divide $a$ or $b$, then it does not appear in any of these factorizations, hence it does not divide $ab$.

5. Feb 24, 2015

### PsychonautQQ

Oh, of course, thank you guys.

Am i doing this next step right?

Keeping the class equation in mind: (|G| = |Z(G)| + [G:N(a)])
"Since p | [G:N(a)] then p | Z(G)|"

We know p| |G| and we know p|[G:N(a)], so p| |Z(G)|.

The way I was thinking of it: if p|b and p|b+c, then p|c.

6. Feb 24, 2015

### BruceW

I am not certain about where you are now in the proof. I'm reading the wiki page on Cauchy's theorem, in proof 1, for general case. Is that where you are too?
The class equation should be: |G| = |Z(G)| + Σ [G:N(a)] Where the sum is over one representative element from each conjugacy class, not in the center of Z. Also, wiki use C(a) instead of N(a), but it doesn't matter, since they are the same when acting on a singleton set, right?
Anyway, they are saying if p divides |Z(G)| then we are done, since the center is Abelian. But if p does not divide |Z(G)|, then we have p divides |Z(G)| + Σ [G:N(a)] but not |Z(G)| . Therefore, there must be at least one term in the sum which is not divisible by p (otherwise this would imply that p does divide |Z(G)| which we have assumed not to be true). So, there must be at least one term where p does not divide [G:N(a)] But p does divide |G|, therefore p must divide |N(a)|. and since 'a' is not central, N(a) is a proper subgroup of G. Finally, they say "This subgroup contains an element of order p by the inductive hypothesis and we are done". So I guess they mean that we use N(a) as our new group, and follow a similar line of reasoning, we either find that p divides |Z(N(a))| or N(a) contains another proper subgroup which is divisible by p, so we can inductively continue.

7. Feb 24, 2015

### jbunniii

Yes, this reasoning is correct. If $p$ divides $b$ then we can write $b = np$ for some integer $n$. If $p$ also divides $b+c$, then $b+c = mp$ for some integer $m$. Therefore, $c = b - mp = np - mp = (n-m)p$, so $p$ divides $c$.