# Finite abelian group

1. Nov 3, 2009

### NanoSaur

Suppose A is a finite abelian group and p is a prime. A^p={a^p : a in A} and A_p={x:x^p=1,x in A}.
How to show A/A^p is isomorphic to A_p.
I tried to define a p-power map between A/A^p and A_p and show this map is isomorphism.
But my idea didnot work right now. Please give me some help.
In addition, How to show the number of subgroups of A of order p equals the number of
subgroups of A of index p.

2. Nov 4, 2009

### ilia1987

There is a theorem: http://en.wikipedia.org/wiki/Finitely_generated_abelian_group" [Broken]
which states that every finite Abelian group can be decomposed into the direct product of groups of prime order (any group of prime order is cyclic btw).
Let's look at the subgroup $$A_p$$. The order of each $$x\in A_p$$ must be p, it's clear that it can't be larger than p, and because p is prime it can't be smaller either. If the order of x is r which is smaller than p, then r must be a divisor of p, but p is prime.
Clearly, every element of A with order p is included in $$A_p$$.
And we have:
$$A_p=P_1 P_2... P_n$$ where every $$P_i$$ is a cyclic group of order p and n is the number of times that these groups occur in the decomposition of A.
Let's look at the subgroup $$A^p$$, if $$A^p\bigcap A_p\neq\{e\}$$ then $$\exists x\in A_p,\ \ x=y^p$$ and then $$\exists y,$$ with order $$p^2$$.
But then let's look at the subgroup generated by various power of y. this is a cyclic group of order $$p^2$$ and as every other abelian group can be decomposed into a product of two order p cyclic groups. so $$y=hg$$ where both h and g have order p. but hg must also have an order p in that case. and so we have that the order of y is p, and thus
$$A^p\bigcap A_p=\{e\}$$ . If we look at the homomorphism:
$$\phi :A \rightarrow A^p \ \ \ \ \phi (a)=a^p$$
the kernel of this homomorphism is clearly $$A_p$$ and we have:
$$im(\phi) \cong A/ker(\phi)\ \ \Rightarrow \ \ A^p \cong A/A_p$$
now we have enough information to conclude that A is decomposed into the product of $$A_P$$ and $$A^p$$ which answers both parts of your question!
During the proof there was ambiguity between semi-direct product and direct product of subgroups but when it comes to abelian groups the two are the same up to isomorphisms.
I am absolutely certain that the proof contains a lot of unnecessary stuff and can be simplified so good luck, and I hope I got the main idea through.

P.S. You can also say that let $$A=P_1 P_2...P_n Q_1 Q2...Q_m...Z_1...Z_l$$ be the decomposition of A into cyclic groups of prime order where each two groups with the same letter have the same order. Then the product of all these groups with the $$P_1...P_n$$ left out should be isomorphic to $$A^p$$, the reason for this being that taking the p-th power of each element is an isomorphism because p is not a divisor of any of the orders of the elements in this subgroup.

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