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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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    $\begingroup$ It does not say that $p$ divides $|H|$, it says that either $p$ divides $|H|$ or $p$ divides $|G/H|$. $\endgroup$
    – Derek Holt
    Commented Apr 23, 2014 at 13:12
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    $\begingroup$ An alternative approach would be to consider the set $\{(a_1,\ldots,a_p) \in G^p: a_1a_2 \cdots a_p = e\}$. $\endgroup$
    – user133281
    Commented Apr 23, 2014 at 13:34
  • $\begingroup$ hi, just a question on this theorem, why do we need the condition of prime? $\endgroup$ Commented Nov 3, 2014 at 16:56
  • $\begingroup$ What is the base case here? $\endgroup$
    – Henry
    Commented Dec 31, 2017 at 22:00

1 Answer 1

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For the alternative approach that @user133281 suggested, see Isaacs' 'Finite Group Theory' exercise 1A.8. Using group actions, it is easier, I think.

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