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Let $|G|$ be an abelian group and let $H = \{g \in G : |g| \text{ divides } 12 \}.$ Prove that $H$ is a subgroup of $G$.

I know that I have to show that $a,b \in H \Rightarrow ab^{-1} \in H$ or $(ab \in H \land a^{-1} \in H).$ But I can't figure out how $|a|$ and $|b|$ dividing $12$ relates to $|ab|$ or $|ab^{-1}|$ dividing $12$.

Oliver G
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2 Answers2

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The crucial observation is that $|g|$ divides $12$ if and only if $g^{12} = e$. With this, it is simple to verify the requirements for subgroup:

  • $e \in H$ because $e^{12}=e$.

  • $a \in H \implies a^{-1} \in H$ because $(a^{-1})^{12}=(a^{12})^{-1}=e^{-1}=e $ (or simply that $|a^{-1}|=|a|$).

  • $a,b \in H \implies ab \in H$ because $(ab)^{12}=a^{12} b^{12}=ee=e$.

The last argument relies crucially on $G$ being abelian.

lhf
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Notice that $|g|$ divides $12$ if and only if $12g = 0$. Because the map $f \colon G \to G$, $g \mapsto 12g$ is a group homomorphism it follows that $H = \ker f$ is a subgroup.

Jendrik Stelzner
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  • I haven't come across the terms "homomorphism" or "ker" in the book I'm reading, is there a way prove this specifically with order relationships? – Oliver G Jul 15 '16 at 12:05