Let $G$ be a finite group such that $H$ is normal in $G$ and $G/H$ has order $7$ then is $G \cong H \times G/H?$

Question

Let $G$ be a finite group such that $H$ is normal in $G$ and $G/H$ has order $7$ then is $G \cong H \times G/H?$

My attempt: Suppose it's true and let $f: G \to H \times G/H$ be the isomorphism. $G$ must have an element $x$ of order $7$ so $f(x) = (h, yH)$ has order $7$ so $h$ must have order $7$ or $h=e$. The former case implies that $49 | |G|$ and the latter case implies that $f(<x>)= \{e\}\times G/H.$ But I haven't been able to reach a contradiction from either of these cases. How should I proceed from here? Thanks in advance.

Answer 1

Take $G=\Bbb{Z}/49\Bbb{Z}$ and $H=7\Bbb{Z}/49\Bbb{Z}$. Then $G/H$ has order $7$ but $G\not\cong H\times G/H$ as $G$ is cyclic.