Is this true for any subgroup of a group?

Question

If $H\leq G$, then is it true that $g_1 g_2 H=g_1 H g_2 H$ for any subgroup $H$, or only if $H$ is normal in $G$? And how is this equality proved?

Otherwise, if the equality above is true for all $H\leq G$, then a function $\pi: G\to G/H$ would be a homomorphism, and thus any subgroup would be normal.

Answer 1

Let $G = S_4$ and $H = \{(1,2),e\}$. Clearly $H \leq G$, yet $H$ is not normal as $(1,3)H = \{(1,3)(1,2),(1,3)\} = \{(1,3,2),(1,3)\}$ while $ H(1,3) = \{(1,2)(1,3),(1,3)\} = \{(1,2,3),(1,3)\}$, which are not equal.

Now $(1,3)(1,2)(1,4)(1,2) = (1,3)(2,4)\in (1,3)H(1,4)H$, however $$ (1,3)(1,4)H = \{(1,3)(1,4)(1,2),(1,3)(1,4)\} = \{(1,3,4,2)(1,3,4)\} $$ which does not contain $(1,3)(2,4)$, showing the claim is false.

Answer 2

\begin{aligned}g_1 g_2 H=g_1 H g_2 H &\iff g_2H=Hg_2H \\ & \iff H = g_2^{-1}Hg_2H \\ &\iff H = H^{g_2^{-1}}H \\ &\iff H^{g_2^{-1}} \leq H \\ &\iff H \leq H^{g_2}\end{aligned}

If $H$ is finite, then this is also equivalent with $H=H^{g_2}$ and thus $g_2\in N_G(H)$. (But in general, it is very well possible that $H\lneqq H^{g_2}$!)

So we observe:

• If $H$ is normal then clearly this always holds.
• If $H$ is not normal, then the equality will sometimes fail. In the finite case, this is obvious since any $g\not\in N_G(H)$ will provide a counterexample $(g_1,g_2)=(1,g)$. In general, we take any element $g\not\in N_G(H)$. Then $(g_1,g_2)=(1,g)$ will be a counterexample, unless it is accidentally the case that $H\lneqq H^{g}$. But then $H^{g^{-1}} \lneqq H$ and then $(g_1,g_2)=(1,g^{-1})$ provides a counterexample.

Answer 3

If for every $g_1, g_2 \in G$ and $H<G$, holds that $g_1g_2H=g_1Hg_2H$ then $H=g^{-1}HgH$ $\forall g \in G$ so $\forall$ $h_1, h_2 \in H$ $\exists$ $h \in H$ such that

$g^{-1}h_1gh_2=h \in$ $H$ $\implies$ $g^{-1}h_1g=hh_2^{-1} \in$ $H$ $\implies$ $g^{-1}hg \in$ $H$ $\forall$ $h \in H$, and so $g^{-1}Hg < H$

This implies that $\forall$ $g\in G$ $gH<Hg$ and $Hg<gH$, so $Hg=gH$ holds ever and $H\unlhd G$.