I have multiple questions regarding: https://math.stackexchange.com/a/864985/997899
Q: Prove that any group of order 15 is cyclic.
A: Let $G$ be a group such that $|G| = 15$.
Show that the group contains normal subgroups of order $3$ and of order $5$. Let's call them $H$ and $K$ respectively.
Prove the following fact: If $H$ and $K$ are normal, $H \cap K = \{e\}$, and $G = HK = \{hk : h \in H, k \in K\}$, then $G \cong H \times K$.
Hints for #2:
- Prove that $H \cap K$ is a subgroup of both $H$ and $K$.
- Show that $HK$ is a subgroup of $G$.
Once you have done these, you are more-or-less finished.
- How to prove: $G=HK=\{hk:h\in H,k\in K\}$? It's mentioned we need to prove first $GK$ is subgroup of $G$ but then what? why this proves it's the same as $G$?
Note: Number of elements in $HK$ isn't $15$ for sure, as different multiplications may result in the same element in the set.
- After I finish 1 why does this solve the problem? How does this say at all that $G$ is cyclic, according to what law?
Best Answer
$HK$ is a group whose subgroups are both $H$ and $K$. Thus, its order is divisible by both $3$ and $5$, i.e. by $15$, which means it is $15$ at least! As it is contained in the group $G$ of order $15$, we must have $G=HK$.
How to then finish the proof? As groups $H$ and $K$ are of prime order, they are cyclic. So $H\cong C_3$ and $K\cong C_5$. Thus, $G=HK\cong H\times K\cong C_3\times C_5\cong C_{15}$.
The last isomorphism above (and generally, if $(m,n)=1$ it is known that $C_m\times C_n\cong C_{mn}$) is one of the equivalent formulations of Chinese Remainder Theorem.