# If $G$ is an abelian group, then $H=\{g\in G \mid |g| \text{ divides }12\}$ is a subgroup of $G$

$$H=\{g\in G \mid |g| \text{ divides }12\}$$

We have to prove that $H$ is a subgroup of $G$.

Consider $a\in G$ and, the group $\langle a \rangle$ where $|a|=12$. This group will contain elements orders of which will be divide 12. This is a cyclic subgroup I think this should be equal to H but I am not sure as I don't know if G is finite.

• If you have $G = \Bbb Z_{12}\times \Bbb Z_{12}$, then $H = G$ but $H$ isn't cyclic. So your idea with $\langle a\rangle$ doesn't work. For another reason it doesn't work, if $G = \Bbb Z_4$, then once again $H = G$, but no element has order $12$. – Arthur Sep 3 '18 at 8:30
• You can not be sure that such an element $a$ exists in general. For all we know, $H$ could be as small as the $1$-element group $\{e\}$. – Suzet Sep 3 '18 at 8:31
• is the number $12$ significant somehow? – Alvin Lepik Sep 3 '18 at 8:32
• How else can I approach this? – Piyush Divyanakar Sep 3 '18 at 8:32
• How would you normally prove a statement like "Show that the subset $A$ of a group $B$ is actually a subgroup"? I would guess that you have a list of three properties, including something very much like "$e\in A$". Try those, and if you can't do them all, come back here with all your results and we'll help you get the rest. – Arthur Sep 3 '18 at 8:34

Hint: We can rewrite $H=\{g\in G\mid 12g=0\}$. Therefore, if you can show that the map $g\mapsto 12g$ is a homomorphism, then you have that $H$ is the kernel of a homomorphism and therefore a subgroup.

• I have just started so I haven't covered homorphisms yet so I couldn't think in this direction, but I understand the idea based on my Linear algebra knowledge. Great solve thanks. – Piyush Divyanakar Sep 4 '18 at 7:21
• @PiyushDivyanakar You can use the rewriting without talking about homomorphisms. Show that if $12g=0$, then $12(-g)=0$, if $12g=12h=0$, then $12(g+h)=0$, and $12(0)=0$. There are more things to check this way, but it is still straight forward. – Aaron Sep 4 '18 at 7:24
• Yes that's how I solved it. I based it on ArsenBerk's answer. – Piyush Divyanakar Sep 4 '18 at 7:26

Instead of seeking a specific example like cyclic groups, you can prove it in a more general way:

HINT: $H \le G$ if and only if

1) $H$ is not empty.

2) $H$ is closed under binary operation of $G$.

3) $H$ is closed under inverses.

• "actually 2) and 3) implies 1) but we need to ensure $H$ is non-empty." Then 2) and 3) do not imply 1). 2) and 3) together with "1') $H$ is non-empty" imply 1). – Arthur Sep 3 '18 at 8:42
• Oh, that is really logical. – ArsenBerk Sep 3 '18 at 8:43
• Yes, we do assume that. It's a convention called vacuous truth, and it often does seem counterintuitive to people when they first encounter it. But it works better with all our other logic than any alternative. – Arthur Sep 3 '18 at 8:48
• Well, after I asked, I thought that it is not counter-intuitive because in an empty set, there is no element that breaks closure under binary operation or closure under inverses. I will check "vacuous truth" out, thank you for that :) – ArsenBerk Sep 3 '18 at 8:50