Without using Lagrange's theorem, show that an abelian group of odd order cannot have an element of even order.

Question

Let $G $ be an abelian group of odd order, say $n$. Then, by fundamental theorem of finite abelian groups, $G$ is a direct product of groups of prime power order. Let $$ G=\mathbb Z_{{p_1}^{n_1}} \oplus \mathbb Z_{{p_2}^{n_2}}\oplus \cdots\oplus\mathbb Z_{{p_k}^{n_k}}.$$ Then $\vert G\vert=n={p_1}^{n_1}{p_2}^{n_2}\cdots{p_k}^{n_k}$. Now let $(g_1,g_2,...,g_k)\in G$. Then $\vert (g_1,g_2,...,g_k )\vert=\mathrm{lcm}(\vert g_1 \vert,\vert g_2 \vert,...,\vert g_k \vert)$.

From here how to move, i don't know but it is given in the hint to make use of fundamental theorem of cyclic groups (but i'm not getting how to apply it here).

Answer 1

Assuming you can prove the fundamental theorem without using Lagrange (which I don't believe), a consequence of the theorem is that, for every $x\in G$, $x^n=1$, because $n$ is a multiple of the order of every cyclic component and, clearly, in a cyclic group $C$ of order $k$, we have $g^k=1$, for every $c\in C$.

The map $x\mapsto x^2$ is an endomorphism of $G$ and, since $n=2k+1$ is odd, we have $$ x=(x^{k+1})^2 $$ so the map is surjective. Thus the map is also injective, therefore $x^2=1$ implies $x=1$ and there's no element of order $2$.

Answer 2

Suppose $g\in G$ has even order, say $2m$, then $g^m$ has order $2$ so assume $g$ has order $2$, then we can partition $G$ into pairs, $\{h,k\}$ where $h=gk$ and $k=gh$.

This is of course impossible since $G$ has odd order, so we have the result by contradiction.