# Prove that any group of order 15 is cyclic?

cyclic-groupsfinite-groupsgroup-theorypermutations

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.

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

$$HK$$ 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.