0
$\begingroup$

I'm working through Abstract Algebra by Beachy and Blair and there is a piece of the puzzle in a proof that I'm missing. The lemma is as follows.

Lemma 7.5.4: Let $G$ be a finite abelian p-group, and $a \in G$ be an element of maximal order. Then, each coset of $\langle a\rangle$ contains an element $d$ such that $\langle d\rangle \cap \langle a\rangle = \{0\}$.

Proof: Let $|a| = p^\alpha$. Given $b \in G$, let $|b| = p^\beta$, and let $p^\gamma$ be the order of $b + \langle a\rangle$ in the factor group $G/\langle a \rangle$. (rest omitted)

In order for $G/\langle a\rangle$ to be a group, $\langle a\rangle$ must be normal in $G$. It is not obvious to me why $\langle a \rangle$ is a normal subgroup of $G$.

I know that because $a$ has maximal order, $\langle a\rangle$ is a maximal cyclic subgroup of $G$. In order for it to not be a normal subgroup, $g\langle a\rangle g^{-1} \neq \langle a \rangle$ for some $g \in G$. So, $g\langle a\rangle g^{-1}$ would have to be a cyclic subgroup of $G$ having the same order as $\langle a \rangle$, so it must contain some other element $a' \neq a$ also having maximal order in $G$. I don't see how this is not possible.

$\endgroup$
  • 1
    $\begingroup$ $G$ is abelian, so any subgroup of $G$ is normal in $G$. $\endgroup$ – Prasun Biswas Mar 23 '18 at 4:22
  • $\begingroup$ Oh... Of course. Thanks for pointing that out, haha! $\endgroup$ – Kevin Hsu Mar 23 '18 at 4:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.