# Proving $G=ST$ if $T$ is a non trivial subgroup of $G$ and $S$ is a subgroup of $H$ where $G$ divides $S$

Let $$G$$ be a finite group and $$S$$ a subgroup of $$H$$ such that $$\mid G \mid/\mid S \mid$$ is a prime number. Assume that T is a non-trivial subgroup of $$G$$ with $$S \neq T$$ and $$ST$$ is a subgroup. Prove that $$G = ST$$.

I have that since $$\mid G \mid/\mid S \mid$$ is a prime number, the gcd of order of $$G$$ and $$S$$ is 1. Then since $$T$$ is a subgoup of $$G$$ where $$T$$ is non-trivial, $$S$$ and $$T$$ should also have order gcd of 1. Hence $$G = ST$$. I think this was a faulty attempt, but I'm lost otherwise.

• $S$ is a subgroup of $G$, thus its order divides $|G|$ and is not coprime to it. – Berci Feb 9 at 1:32
• Anyway, it's not true as written, since $T$ might be a proper subgroup of $S$. – Berci Feb 9 at 1:34

As written, this problem is wrong. The condition $$S\not=T$$ is not enough (suppose that $$T$$ is any nontrivial subgroup of $$S$$). For an example, let $$G=D_4=\langle\sigma,\tau:\sigma^4=\tau^2=e,\sigma\tau=\tau\sigma^{-1}\rangle$$, $$S=\langle\sigma\rangle$$ and $$T=\langle\sigma^2\rangle$$. In this case, $$[G:S]=2$$, $$T$$ is not trivial and $$T\not=S$$, but $$ST=S\not=G$$.

You need the slightly stronger condition that $$T\not\subseteq S$$. In this case, you have the following:

You know that $$S\varsubsetneq ST\subseteq G$$. To complete the proof, you only need the following three facts:

1) $$[G:S]$$ is prime,

2) $$[G:S]=[G:ST][ST:S]$$, and

3) $$[ST:S]>1$$.

Now, what are the possibilities for $$[ST:S]$$?