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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.

sequence
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    The two (left) cosets you've written are always equal. Clearly not all subgroups are normal, so your conclusion is false. Perhaps you made a typo? – Yeldarbskich Nov 27 '15 at 04:52
  • Sorry, I've now made a correction. The identity in question is $g_1 g_2 H=g_1 H g_2 H$. – sequence Nov 27 '15 at 04:59
  • Have you tried working this out yourself? It's not that hard. If you do feel free to self-answer. – djechlin Nov 27 '15 at 05:04
  • I'm not sure what "this" means in "if this is not true" since you make conflicting claims right before it. – djechlin Nov 27 '15 at 05:06
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    As $H$ is a subgroup, it contains $e_G$, therefore every $g_1g_2h\in g_1g_2H$ satisfies $g_1g_2h = g_1eg_2h \in g_1 H g_2 H$, so we have $g_1g_2H \subseteq g_1 H g_2 H$. If you want to show that the claimed equality is not true, pick a small group, with a small non-normal subgroup, and try to find a counter-example. – James Nov 27 '15 at 05:06
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    But if this equality is always true then every subgroup would be normal, since if $\pi: G\to G/H$ is a homomorphism, it's kernel is $H$. Something is wrong somewhere. – sequence Nov 27 '15 at 05:21
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    Related (though not really a duplicate): http://math.stackexchange.com/questions/14282/why-do-we-define-quotient-groups-for-normal-subgroups-only/ – Tobias Kildetoft Nov 27 '15 at 10:22

3 Answers3

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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.

James
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\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.
Myself
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  • What is $H^{g_2^{-1}}$? I'm not sure what this notation means. – sequence Dec 01 '15 at 02:12
  • It is a notation for conjugation: $x^y = y^{-1}xy$ for any two elements $x$ and $y$ and $H^g=g^{-1}Hg$ whenever $H$ is a subset (in particular if $H$ is a subgroup). – Myself Dec 01 '15 at 13:24
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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 $H\unlhd G$.

Esteban G.
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  • It is not always true that $g^{-1}Hg\leq H$ implies that $H=g^{-1}Hg$ (there are counterexamples when $H$ is infinite). – Myself Nov 27 '15 at 17:54
  • I've been thinking and what you say sound possible but I can't find a counterexample. Could you help with that?? – Esteban G. Nov 27 '15 at 18:21
  • What about now? – Esteban G. Nov 27 '15 at 18:42
  • A counterexample (the standard counterexample) is given by the subgroup $H = \langle \begin{pmatrix}1&1\0&1\end{pmatrix}\rangle \leq\mathsf{SL}_2(\mathbb Q)$ and the element $g=\begin{pmatrix}2&0\0&1/2\end{pmatrix}$ – Myself Nov 27 '15 at 20:17
  • I think your editted answer is correct btw. (If you read $<$ as $\leq$.) – Myself Nov 27 '15 at 20:25