For a normal subgroup $H$ of $G$. how to prove $XH=HX$ for any part of $G$ ? Is union distributive over multiplication?

Question

I am trying to prove that if $H$ is a normal subgroup of $G$ (i.e. $xH=Hx, \forall x\in G$) then for any subset $X$ of $G$ we have $XH= HX$. The hint given by the Wikipedia page is "take the union". Following this hint we would do something like this:

$$x_1 H = Hx_1$$ $$...$$ $$ x_n H = H x_n$$ $$...$$ $$\implies x_1 H \cup ... \cup x_n H \cup ... = Hx_1 \cup ... \cup Hx_n \cup ... $$

But then, I don't how we can prove that we can factor by $H$ the LHS and RHS. I am unsure of the distributivity of the union over multiplication.

I also tried another way:

Let's say, to simplify, that $H$ has only two elements $h_1$ and $h_2$.

Then, we would not necessarily have a correspondance between $h_1$ and itself (and between $h_2$ and itself) when taking the conjugation. We could have for instance (let $x\in G$):

$$x h_1 = h_2 x \tag{1}$$ $$ xh_2 = h_1 x \tag{2}$$

These equalities are valid for any $x$. So let $x' \in G$. We have:

$$x' h_1 = h_2 x' \tag{1'}$$ $$ x'h_2 = h_1 x' \tag{2'}$$

From there, I have tried to multiply $(1)$ with $(1')$, join $(1)$ and $(1')$ with logical disjunction but could not get to the desired result.

Any help would be appreciated,

Many thanks.

Answer 1

This is similar to your second attempt. Try proving $HX\subset XH$ and $XH\subset HX$.

Hint:

Let $a\in XH\implies a=xh$ for some $x\in X,~h\in H.$
Since $xH=Hx$, the element $xh\in xH\implies xh\in Hx\implies xh=h_1x$ for some $h_1\in H$.
Hence, $a=xh=h_1x\in HX$.

Answer 2

If $X=\varnothing$, then $XH=\varnothing =HX$. Suppose $X\neq \varnothing$. Let $g\in XH$. Then $g=xh$ for some $x\in X, h\in H$. But $H\stackrel{(1)}{\unlhd}G$, so, in particular, $g\in xH\stackrel{(1)}{=}Hx\subseteq HX$. Thus $XH\subseteq HX$. Similarly, $HX\subseteq XH$.