Prove that if $H$ is a subgroup of group $G$, $H\circ x=H$ if and only if $x\in H$

Prove that if $$H$$ is a subgroup of group $$G$$, $$H\circ x=H$$ if and only if $$x\in H$$.

Not sure how to start this proof about the right coset.

• Avoid no-clue question: See How to ask a good question. – Saad Apr 22 '20 at 12:14
• If $H\cdot x=H$ there is a $y\in H$ such that $yx=1$ – Dan Sheppard Apr 22 '20 at 12:37
• What have you tried? For $\implies$, just split into cases according to $x \in H, x \not \in H$. $\impliedby$ is even easier. – Physical Mathematics Apr 22 '20 at 12:43

To prove this, you really just need to apply the definitions of a group and also you need to understand that a group is closed under its operation.

We can prove a general one:

Let $$H$$ be a subgroup of $$G$$, $$a,\,b\in G$$. Then $$aH=bH$$ if and only if $$a^{-1}b\in H$$.

If $$a^{-1}b\in H$$, then $$b=ah_0$$ for some $$h_0\in H$$; thus $$bh=a(h_0h)\in aH$$ and $$ah=b(h_0^{-1}h)\in bH$$ for any $$h\in H$$, which means $$aH=bH$$.

If $$aH=bH$$, then $$ah_1=bh_2$$ for some $$h_1,\,h_2\in H$$, and so $$a^{-1}b=h_1h_2^{-1}\in H$$.

Forward implication:

\begin{alignat}{1} &H=Hx \Longrightarrow \\ &H\subseteq Hx \Longrightarrow \\ &\forall h \in H, \exists h'\in H \mid h=h'x \Longrightarrow \\ &\exists h'\in H \mid e=h'x \Longrightarrow \\ &H\ni h'=x^{-1}\Longrightarrow \\ &x\in H \\ \tag 1 \end{alignat}

Reverse implication:

\begin{alignat}{1} &x\in H \Longrightarrow \\ &hx\in H, \forall h\in H\Longrightarrow \\ &Hx\subseteq H \\ \tag {2a} \end{alignat}

and:

\begin{alignat}{1} &x \in H \Longrightarrow \\ &\forall h \in H, h=(hx^{-1})x \in Hx \Longrightarrow \\ &H\subseteq Hx \\ \tag {2b} \end{alignat}

By $$(2a)$$ and $$(2b)$$, $$x\in H \Longrightarrow H=Hx$$.