Given a non-commutative group of order $\mathrm{ord}(G)=343$. Prove that $G$ is a simple group.
So I have to show that the only normal subgroups are the trivial group and the group itself. But I don't know how to use the order of the group to show it. The non-abelian part is also not helping really.
In fact $G$ is not simple! This holds in general. Because $|G|=p^n$ where $p$ is a prime and $n>1$ we have that $Z(G)\neq 1$. If $G$ is not abelian then $Z(G)\neq G$ and thus $G$ is not simple because $Z(G)$ is normal in $G$.
In case now $G$ is abelian. Then $G=Z(G)$ and every subgroup of $G$ is normal. Let $g\neq 1$. If $\langle g \rangle \neq G$ then we are done because we found a non-trivial subgroup of $G$.
If $G=\langle g \rangle$ then $G$ is cyclic and let $H= \langle g^p \rangle$. We have that $1\neq H\lhd G$ because :
If $H=1$ then $g^p=1 \Rightarrow|G|=|\langle g \rangle |=p$ -contradiction.
If $G=H$ then $g\in \langle g^p \rangle \Rightarrow g=g^{kp} \Rightarrow g^{kp-1}=1 \Rightarrow o(g) \mid kp-1 \Rightarrow p \mid kp-1 \Rightarrow p=1$ again a contradiction.