Clarification on finding another subgroup given the order of two existing subgroups

Question

If we assume that G is abelian and that it has a subgroup of order 7 and another of order 11. If we were asked to find another subgroup of this group would we take the least common multiple of the two orders? In this case, 77? If not, what theorem do we use to figure this out and what role does G being abelian play?

Answer 1

Yes you are right. Given that G is abelian and it has a subgroup of order 7 and 11.As 7 and 11 are prime numbers these subgroups are cyclic.Thus we have a elements of order 7 and 11.As the group is abelian so there will be an elements of order lcm(7,11)=77(see proof of this theorem from Dummit and Foote).Thus you have a subgroup of order 77.

Answer 2

If $a$ is a non-identity element of the subgroup of order seven, call it $H_1$, then $a$ is itself of order seven ($7$ is prime so Lagrange's theorem excludes all other possibilities). Similarly we can find an element $b\in H_2$ of order eleven from the other subgroup. The key observation is that $H_1\cap H_2$ is necessarily the trivial subgroup, again by Lagrange.

The question we can ask ourselves is: What's the order of the product $ab$? Because $G$ is abelian we have $(ab)^n=(ab)(ab)\cdots(ab)=a^nb^n$ for all positive integers $n$. If $(ab)^n=a^nb^n=1_G$, then we get that $a^n=x=b^{-n}$. But here $a^n\in H_1$ and $b^{-n}\in H_2$. Therefore $x\in H_1\cap H_2$, so by our earlier considerations $x=1_G$. By basic properties of the order of an element the equation $a^n=1$ implies that $7\mid n$. Similarly $b^{-n}=1$ implies that $11\mid n$. Hence we can conclude that $(ab)^n=1_G$ iff $77\mid n$. This means that the element $ab$ generates a cyclic subgroup of order $77$.

Note that $G$ being abelian was absolutely essential here. For example in the group $S_{11}$ we could have $a=(1234567)$ and $b=(123456789AB)$ (I use $A=10$ and $B=11$ to avoid confusion in the cycle notation). In that case we have $$ ab=(135789AB246), $$ another element of order $11$, not $77$.