Prove that any group of order 15 is cyclic?

Question

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 $|G| = 15$.

1. Show that the group contains normal subgroups of order $3$ and of order $5$. Let's call them $H$ and $K$ respectively.

2. 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.

1. 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.

2. After I finish 1 why does this solve the problem? How does this say at all that $G$ is cyclic, according to what law?

Answer 1

$HK$ 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.

Answer 2

Cauchy's theorem (group theory) : Let, $G $ is s a finite group and $p$ is a prime number dividing the order of $G$ (the number of elements in G), then $ G$ contains an element of order $p$.

$G$ be an abelian group . Let $a,b$ be elements of $ G$ with order m and n, respectively. If $m$ and $n$are relatively prime, then show that the order of the element $ab$ is $ mn$.

Hence, $G$ contains an element of order $3$ say $a$ and an element of order $5$ say $b$. Then order of $ab$ is $3×5=15$

Since, $G$ is a group of order $15$ and contains an element $ab$ of order $15$.

Hence, $G=<ab $

Sylow's theorem is a another good options.

But an important corollary, if $order(G) =pq$ ,where$ p, q$ are primes with $p<q$ and $p $ doesn't divides $q-1$ , the the group $G$ is cyclic.

Here, $order(G)=3•5 $ and $3$ doesn't divide $( 5-1)=4$