Clarification on Lagrange's Theorem and the group $Z_n$ of integers modulo $n$

Question

I want to show that $Z_3 \leq Z_6 \leq Z_{12}$.

Since $|Z_3|$ and $|Z_6|$ divide $|Z_{12}|$, then they must be subgroups of $Z_{12}$. Similarly, $Z_3 $ is a subgroup of $Z_6$. I feel like here I am using the converse of Lagrange's Theorem. Does the converse hold? If not, what other easy way can we use to show that $Z_3 \leq Z_6 \leq Z_{12}$?

So just to see if I have gotten Lagrange's Theorem: I can only use Lagrange's theorem if we know, for example, that $H$ is a subgroup of $G$. In other words, if we already know that $H \leq G$, then it must be the case that $|H|$ divides $|G|$. Correct? Thanks.

Answer 1

You're correct. The statement of Lagrange's Theorem is that IF $H\leq G$ is a subgroup and $G$ is finite, then $|H|$ divides $|G|$. The converse of Lagrange's Theorem is in general not true, but there are partial converses of this, namely the Sylow Theorems/Cauchy's Theorem.

A counterexample to this is given by $A_4\leq S_4$, the alternating group. The order of $A_4$ is $4!/2=12$, but there is no subgroups of order $6$.

Answer 2

The converse does not hold. For example, $|Z_4| = 4$ divides $|Z_2 \times Z_2| = 4$ (and vice versa) but neither of these groups are subgroups of the other, as this would force the groups to be isomorphic, which they are not.

Hint On the other hand, the groups $Z_p$ are each generated by a single element $g$, (for example, we can always take $g = [1]$). In our case $p = 12$; for any generator $g$, what can we say about the group generated by $g^2$?