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
 A: 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.
A: 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.
A: 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.
