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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.

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    $\begingroup$ What do you think of $G= \mathbb{Z}/49\mathbb{Z}$ ? $\endgroup$ – Maxime Ramzi Apr 19 '18 at 20:22
  • $\begingroup$ Let us consider a commutative group $G$ with $49$ elements... The theorem of invariant factors gives the possibilities $\mathbb Z/49$ and $\mathbb Z/7\times \mathbb Z/7$. Starting with $G$ with an element of order $49$... $\endgroup$ – dan_fulea Apr 19 '18 at 20:24
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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.

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  • $\begingroup$ You're right my mistake $\endgroup$ – Mike Apr 19 '18 at 20:33

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