Let $G$ be an abelian group of order 72. How many subgroups of order 8 and 4 can it have?
I have listed all possible abelian groups there are 6. Then i said that if I'm looking for an abelian group of order 8 I have 3 possible abelian groups and for each 1 of the 6 first abelian groups must check if 1 of the 3 of the later ones are subgroups. But that is taking forever and I don't have any easy way to check if they are subgroups. I use "If A is a subgroup of G and B is a subgroup of H, then the direct product A × B is a subgroup of G × H". So I found that for every possible G there is 1 subgroup of order 8. But is that correct? If I have a subgroup of a direct product say A × B of a G SxS1xS2x…xSn then A must be subgroup of S and B of S1?
Best Answer
Hint: If $|G|=72$ and the subgroup $|H|=8$, then $[G:H]$ is an odd number. Thus $H$ is a Sylow $2$-subgroup of $G$. Those are known to all be conjugates of each other.
Of course, if you have not seen Sylow theory yet (a live possibility), then this was not very helpful to you. In that case, try and prove that $$H=\{x\in G\mid 8x=0\}$$ is a subgroup. Because it contains all the elements of $G$ with order that is a power of two, it is unique. It is a bit harder to show that $H$ has eight elements, but that does follows from the structure theory of f.g. abelian groups.
If you have not seen the structure theorem of f.g. abelian groups (another live possibility), then do the following. Prove that $G=H\oplus K$, where $K$ consists of elements $y\in G$ such that $9y=0$. A hint for that: for all $x\in G$ we have $x=9x-8x$. Show that $9x\in H$, $8x\in K$.