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> [itex]\bigcup[/itex] {e} is a subgroup of G. If so, then since G = < <x> [itex]\bigcup[/itex] A>, and <x> [itex]\bigcap[/itex] A = <e>, that G = <x> [itex]\oplus[/itex] A.

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

I feel the key is that I must show that if a,b [itex]\in[/itex] G, and [itex]\exists[/itex]x such that ab [itex]\in[/itex] <x>, then [itex]\exists[/itex]y[itex]\in[/itex]G such that a,b [itex]\in[/itex] <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: [itex](G\setminus <x>) \cup \{e\}[/itex] is not a subgroup.

 

[TopCat]

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