# Find a direct summand of a finite abelian group

1. Jul 20, 2012

### TopCat

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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 $\exists$x such that ab $\in$ <x>, then $\exists$y$\in$G 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.

2. Jul 20, 2012

### micromass

Staff Emeritus
I think that what you're trying to prove is false. That is: $(G\setminus <x>) \cup \{e\}$ is not a subgroup.

3. Jul 20, 2012

### 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 $\cong$ $\sum$ $<e> \oplus \textbf{Z}_{p_{i}^{n_i}}$, with $i \geq 2$, which is a subgroup?

Last edited: Jul 20, 2012