Find a direct summand of a finite abelian group

[TopCat]

Homework Statement

If G is a finite abelian group, and x is an element of maximal order, then <x> is a direct summand of G.

Homework Equations

The Attempt at a Solution

I claim that the hypothesis implies that A = G\<x> \bigcup {e} is a subgroup of G. If so, then since G = < <x> \bigcup A>, and <x> \bigcap A = <e>, that G = <x> \oplus A.

Pf of claim: A is obviously associative, has an identity by definition, and since <x> is a group, b \in A \Rightarrow b^-1 \in A. I'm struggling to show that A is closed.

I feel the key is that I must show that if a,b \in G, and \existsx such that ab \in <x>, then \existsy\inG such that a,b \in <y>. Then the maximality of |x| will require y=x, whence A is closed. But I need a nudge here. I just can't get a rigorous proof of this.

 

[micromass]

I think that what you're trying to prove is false. That is: (G\setminus &lt;x&gt;) \cup \{e\} is not a subgroup.

 

[TopCat]

Since G is finite abelian, G \cong \sum \textbf{Z}_{p_{i}^{n_i}}. If |x| has maximal order, then isn't A just removing one of the direct summands, so that, say, A \cong \sum &lt;e&gt; \oplus \textbf{Z}_{p_{i}^{n_i}}, with i \geq 2, which is a subgroup?