If $G$ is an Abelian group and contains cyclic subgroups of order 4 and 6, what other sizes of cyclic group G must contain?

Question

Let $G$ be an abelian group. Let $a,b \in G$ such that $|a|=4, |b|=6$.

$(ab)^{24}=a^{24}b^{24}=e$, identity of G. Hence $|ab|$ can be $1,2,3,4,6,8,12,24$

If $|ab|=8$ then $(ab)^8=e\implies e=a^8b^8=b^2$, which is a contradiction and hence $|ab|\ne 8,4,2,1$

If $|ab|=6$ then as above, $a^2=e$, which is again a contradiction and hence $|ab|\ne 6,3$

Clearly, $(ab)^{24}=e$ but we also have $(ab)^{12}=e$ and hence $|ab|=12$
Therefore, by fundamental theorem, subgroups of cyclic group $\langle ab\rangle $ can be of order $1,2,3,4,6,12 \;$(all divisors of 12).

Is my solution correct? Or do I have to eliminate some more numbers from $1,2,3,4,6,12$? If yes, then how should I proceed? Thanks for your time. :)

Answer 1

What you did is good. Remain to be proven that none of the orders that you found can be eliminated.

Clearly $1, 2, 3, 4, 6$ can't be eliminated:

• Identity element is of order $1$.
• If $\vert a \vert =4$ then $\vert a^2 \vert = 2$.
• $a$ is of order $4$.
• If $\vert b \vert = 6$ then $\vert b^2 \vert =3$.
• $\vert b \vert$ is having $6$ as order.

Remains the question of the existence of an element of order $12$... which is the case of $a b^2$. Can you prove it?