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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.

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

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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 $a^{-1}b=h_1h_2^{-1}\in H$.

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

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