Mathematics Asked by conchild on December 10, 2020

Please validate the following proof.

Let $H$ be a subgroup of the Abelian group $G$ with elements $h in H$ and $g in G$ but $g notin H$.

**Theorem**

The product $gh notin H$.

**Proof**

Let $x = gh$. Assume $x in H$, then by multiplying by $h^{-1} in H$ we have that $xh^{-1} = gh(h^{-1}) = g(h h^{-1}) = g$. So we have composed two elements of $H$ and obtained an element $g notin H$ and this violates the closure of the subgroup, hence $gh notin H$. $square$

Hypothesis: $Hleq G, gin G-H, hin H$.

Suppose that $ghin H$ and we will have a contradiction/

Since $hin H$ and $H$ is a subgroup we have $h^{-1}in H$.

Since $h^{-1}, ghin H Rightarrow (gh)cdot h^{-1}=gin H$ contradiction.

I used that fact that "if $a,bin Hleq G$ then $ab,bain H$"

Correct answer by 1123581321 on December 10, 2020

In fact

If $H$ is a subgroup of $G$ and $h in H$, then for all $g in G$ we have $gh in H iff g in H$.

Indeed, $gh in H implies g in Hh^{-1} subseteq H$. Conversely, $g in H implies gh in H$. Both claims follow because $H$ is a subgroup and so is closed under products and inverses.

Answered by lhf on December 10, 2020

Yes. Or you could say you have shown $gin H$. Either way, contradiction. That completes the proof.

Answered by Chris Custer on December 10, 2020

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