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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.

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

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