# Corollaries of the Theorem of Finite Abelian Groups

1. Aug 13, 2011

### Barre

1. The problem statement, all variables and given/known data
If G is a finite abelian group and p is a prime such that $p^n$ divides order of G, then prove that G has a subgroup of order $p^n$

2. Relevant equations
Theorem of Finite Abelian Groups: Every finite abelian group G is a direct sum of cyclic groups, each of prime power order.

3. The attempt at a solution
I of course have worked on a solution myself, and it's simply the last step that I'm unsure about.

Since G is finite, by Fundamental Theorem of Abelian Groups we know that

$G = C_1 \times C_2 \times ... \times C_t$

for a finite number of cyclic groups $C_i$ of prime power order. Also, order of G and it's direct summands share the following relation:
$|G| = |C_1||C_2|...|C_t|$, and since $p^n$ divides |G|, there must be a finite set of summands such that product of their orders is equal to exactly $p^n$. Let $C_{p_1}, C_{p_2}, ... ,C_{p_m}$ be those summands. The set:
$S = \{(e_1, e_2,...,a_{p_1},...,a_{p_2},...,a_{p_m},..., e_t) | a_{p_i} \in C_{p_i} \}$,
where all $e_i$ are the identity elements, is clearly a subgroup of $C_1 \times C_2 \times ... \times C_n$, of order $|C_{p_1}||C_{p_2}| ... |C_{p_m}| = p^n$.

Is it now OK to assume that since $G = C_1 \times C_2 \times ... \times C_t$, that G has a subgroup of order $p^n$ isomorphic to the subgroup S ?

It seems logical to me that this is the case, but I need confirmation. Also, is there a more obvious way of proving it? Of course all of this is just a consequence of Sylow theorems, but these come in the next section, so all I want to use is very elementary group theory + the theorem I described.

I'm also sorry about the set S, it might be totally unreadable but I had no idea how to otherwise describe it :D

2. Aug 13, 2011

### micromass

Seems ok!

3. Aug 13, 2011

### Barre

Actually when I think more about this, it seems like there is a special case in which this does not work.

Assume $G = Z_8$. This group is also a product of cyclic abelian groups, namely itself. Order is clearly 8, and 4 (2^2) divides the order, but there is no summand of degree 4. Obviously $Z_8$ contains a subgroup of order 4 anyway, so the proof might not be all lost.

edit: if a summand group $C_i$ has order n then subgroups of any order dividing n can be constructed, as if a generates the group, |a| = n and n = dm then $({a^d})^m = 0$, and the element $a^d$ generates a group of order m. I think this would help cover up the part I missed + probably prove in general the converse of Lagrange's Theorem for abelian groups.

Last edited: Aug 13, 2011
4. Aug 16, 2011

### Barre

Shameless bump and also a new idea on how to approach this one. I've still not figured out a 'nice' proof of this, where my definition of 'nice' is simple (since the fact to be proven is IMO intuitive).

My new try has been by induction on the exponent of divisor $p^n$. I use Cauchy's Theorem for Abelian case which basically is a corollary of Theorem of Finite Abelian Groups. Note that I want to prove this without Sylow's theorems.

$n = 1$ it is clearly true, as Cauchy's Theorem for Abelian Groups says that if order of G is divisible by p, then it contains an element a of order p which clearly generates a cyclic group $<a>$ of order p. Inductive assumption will then of course be that if order of G is divisible by $p^{n-1}$, then G contains a subgroup of this order.

Assume order of G is divisible by $p^n$. Then it certainly also is divisible by $p^{n-1}$, so it contains a nontrivial subgroup of order $p^{n-1}$. Call this subgroup for S. Then, since G is abelian, quotient group $G/S$ certainly exists, and we have relation of orders of groups:
$|G| = |G/S||S|$
Since order of G is divisible by $p^n$, then it follows from the relation that p divides $|G/S|$ and by Cauchy's Theorem, G/S contains an element of order p. Could I somehow 'construct' a group of order $p^n$ by direct product of the S and the group generated by element of order P in G/S?

I'm pretty much stuck. Is there a way to go on, or have I hit a wall?
If somebody knows, could you point me to a place on the internet where this or similar proof has been written out?