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### Problem 8

Let $G$ be an abelian group and let $H=\{ g \in G \mid \ \mid g \mid$ divides $12 \}$. Prove that $H$ is a subgroup of $G$. Would your proof be valid if the number $12$ were replaced by another number? State the general result.

**Solution**

Let $G$ be an abelian group and let $H=\{ g \in G \mid \ \mid g \mid$ divides $12 \}$.

(i)

Clearly, $H \subseteq G$, since $H$ is defined to contain only elements that are in G.

(ii)

Let $a,b \in H$.

then $\mid a \mid = m, \mid b \mid = n$.

hence, there exists $k, l \in \mathbb{Z} ^+$ such that $km=12$ and $ln=12$, by definition of "divides."

Now, consider $(a*b)^{12}.$

Observe that, since $\mid a \mid$ divides $12$ and $\mid b \mid$ divides $12$, $a^{12} = e$ and $b^{12} =e$.

Now, consider $a^{12}*b^{12} = e*e = e$.

Thus, $(a*b)^{12} = a^{12} * b^{12} =e$ since $G$ is abelian.

Since $(a*b)^{12} = e$, the order of $a*b$ is divisible by $12$.

Therefore, $a*b \in H$, so $H$ is closed under $*$.

(iii)

Observe that $e \in G$, since $G$ is a group.

Note, $\langle e \rangle = \{ e \}$, so $\mid e \mid =1$.

Since $1$ divides $12$, $e \in H$.

Now, since $e$ is the identity for all $a \in G$ and $H \subseteq G, e$ is the identity for $H$.

(iv)

Let $a \in H$.

Then, by definition of $H, \exists a^{-1} \in H$ such that $aa^{-1}=a^{-1}a=e$. Since $a\in H$, $|a|$ divides 12, so $a^{12}=e$.

Observe that $(a^{-1})^n=a^{-n}=(a^n)^{-1}=e$, so $|a^{-1}|$ must divide 12.

Therefore, the inverse of $a$ is in $H$ for all $a \in H$.

All conditions hold, so $H$ is a subgroup of $G$.

$\hspace{250pt} \clubsuit$

This would hold even if $12$ were replaced by another element of $\mathbb{Z}^+$.