Problem
Find the number of subgroup of order 3 in an abelian group of order 24?
Attempt Let $n_3$ be the number of subgroup of order 3 . Then by Sylow's theorem $n_3|8$ and $n_3 \equiv 1mod 3$. From this we can tell that $n_3$ can be 1,4 or 8. Is this correct?
Let $G$ be an abelian group with $|G|=24=2^3\times3$ and $H$ be a $3$-Sylow subgroup. Let $n$ be the number of $3$-Sylow subgroups. Then by the Sylow theorem we know $n\equiv 1\mod 3$ and $n|[G:H]$ i.e $n|8$. Thus $n$ can be $1,2,4$ or $8$, but $2$ and $8$ are not congruent to $1\mod 3$ so we can rule those out.
Now all $3$-Sylow subgroups are conjugate and in an abelian group every subgroup is normal, it thus follows that there is exactly one subgroup of order $3$.