# Let $G$ be a finite abelian group and let $p$ be a prime that divides order of $G$. then $G$ has an element of order $p$

Let $G$ be a finite abelian group and let $p$ be a prime that divides order of $G$. then $G$ has an element of order $p$

Proof When $G$ is abelian. First note that if $|G|$ is prime, then $G \approx Z_p$ and we are done.

In general, we work by induction. If $G$ has no nontrivial proper subgroups, it must be a prime cyclic group, the case we’ve already handled.

So we can suppose there is a nontrivial subgroup $H \leq G$. Either $p$ divides $|H|$ or $p$ divides $|G/H|$.

In the ﬁrst case, by induction, $H$ has an element of order $p$ which is also of order $p$ in $G$ so we’re done.

In the second case, if $g H$ has order $p$ in $G/H$ then $|g H|$ divides $|g|$, so $\langle g \rangle \approx Z_{kp}$ for some $k$, and $kg$ ∈ G has order $p$.

I am confused over how

(i) Either $p$ can divide $|H|?$

(ii) In the second case, if $g H$ has order $p$ in $G/H$ then $|g H|$ divides $|g|$, so $\langle g \rangle \approx Z_{kp}$ for some $k$, and $kg$ ∈ G has order $p$

$p$ was supposed to divide $|G|$ . but why does the statement say that $p |~~ ||H|$

Help will be really appreciated. Thanks

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It does not say that $p$ divides $|H|$, it says that either $p$ divides $|H|$ or $p$ divides $|G/H|$. –  Derek Holt Apr 23 '14 at 13:12
An alternative approach would be to consider the set $\{(a_1,\ldots,a_p) \in G^p: a_1a_2 \cdots a_p = e\}$. –  user133281 Apr 23 '14 at 13:34
hi, just a question on this theorem, why do we need the condition of prime? –  Aha Nov 3 '14 at 16:56