Order of abelian group is even

Let $(G, +)$ be a finite abelian group. Suppose $\exists x\in G$ such that $x + x = \textbf{0}$ (where $\textbf{0}$ is the neutral element in G). Show that if $x \neq \textbf{0}$ then $|G|$ is even.

I'm assuming the best way is to show the construction of $G$ as the union of sets $\{a, a+x\}$ in which case the order is automatically even. Does that make sense? Any pointers would be appreciated.

Thanks

• This has 3 upvotes, a star and is now a hot question. Guys this is a classic first year homework question, a long with "prove that every group of prime order is cyclic" and co. It also shows no research because I'm pretty sure I asked this once here! Not saying DV, just move on and/or read! Sep 28, 2015 at 23:32

Since $x+x = 0$, the set $H := \{0,x\}$ is a subgroup of $G$ of order $2$. Lagrange's Theorem implies that $|G| = |G:H||H| = 2 |G:H|$, so $|G|$ is even.

• We haven't seen Lagrange's Theorem yet, but I'll look into it and see if I can come up with an answer from it without using it directly. Thanks!
– mast
Sep 28, 2015 at 22:39
• The proof of it should give you a good idea. Sep 28, 2015 at 22:40

Your idea is right on target!

Let $C_a = \{a, a+x\}$ for $a\in G$. Prove that when $a$ varies over $G$, the $C_a$ form a partition of $G$: their union is all of $G$ and $C_a$ and $C_b$ are either disjoint or equal. Finally, prove that $C_a$ has exactly two elements.

The fact that $G$ is abelian is immaterial.

This is the proof of Lagrange's Theorem for this special case.

We know that for $x \in G$ we must have $x^{|G|} = 0$, therefore if $x$ has order $2$ and $|G|$ was odd, then $x^{|G|} = x^{2k+1} = x$ (for some $k$). As we assumed $x \neq 0$ this proves that $|G|$ is even.

EDIT: If we do not want to use Lagrange's theorem, we can combine the answer linked in the comments and my answer to get the following:

Let $G = \{ x_1, \ldots, x_n \}$. We want to show that $n$ is even. Let us for the sake of simplicity say $x_1 + x_1 = 0$ and $x_1 \neq 0$. Now $G = x_1+G = \{ x_1+x_1, \ldots, x_1+x_n \}$, so $$x_1 + x_2 + \ldots + x_n = (x_1 + x_1) + (x_1 + x_2) + \dots (x_1 + x_n)$$ therefore $x_1 + x_1 +\ldots + x_1 = 0$ (where this sums $x_1$ $n$-times). Now we use the argument above.

• You need Lagrange's Theorem for this argument, unless you have an answer for my pet question
– lhf
Sep 28, 2015 at 22:44
• Ah, I did not realize that when I posted it, I will think of another proof that does not rely on Lagrange's Theorem. Sep 28, 2015 at 22:46
• Although, in this abelian case, one can use this answer Sep 28, 2015 at 22:51