# Subgroup element composition

Mathematics Asked by conchild on December 10, 2020

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