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

$G$ is an Abelian group of odd order.

By the FTFGAG,

$G \cong \mathbb{Z_{p_{1}^{r_{1}}}} \times \mathbb{Z_{p_{2}^{r_{2}}}} \times ... \times \mathbb{Z_{p_{i}^{r_{i}}}}$.

$|G| = p_{1}^{n_{1}} \cdot p_{2}^{n_{2}} \cdot ... \cdot p_{i}^{n_{i}}; p_{i}\neq 2$

$(a_{1},a_{2},...,a_{n}) \in G; a_{i} \in \mathbb{Z_{p_{i}}}$

$|(a_{1},a_{2},...,a_{n})| = lcm(|a_{1}|,|a_{2}|,...,|a_{n}|)$.

Since $\mathbb{Z_{p_{i}^{r_{i}}}}$ is cyclic with cyclic subgroups, $|a_{1}| \in \mathbb{Z}$ has to divide $|\mathbb{Z_{p_{i}^{r_{i}}}}| = p_{i}^{r_{i}}$. Since $p \neq 2$, $|a_{1}|$ must be odd, thus making the $lcm$ odd as well.