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Let $G$ be a group and $H$ be a not normal subgroup of $G$. I need to prove that there are two left cosets of $H$: $g_1H,g_2H$ such that $g_1Hg_2H$ is not a left coset of $H$.

I tried to assume the contrary and get to that $H$ is normal, which will mean a contradiction. I tried just using definitions or to find an homomorphism which I can apply the first isomorphism theorem with maybe. However, it didn't work out for me.

Any help would be appreciated.

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  • $\begingroup$ Try proving the contrapositive; that is, the product is a left coast for all pairs implies $H$ is normal. $\endgroup$
    – Shaun
    Apr 2, 2019 at 16:04
  • $\begingroup$ My previous comment isn't very helpful, considering that you've tried a proof by contradiction. Sorry. $\endgroup$
    – Shaun
    Apr 2, 2019 at 16:10

2 Answers 2

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Suppose that the product of every left cosets is a left coset. This implies that for every $g$, $gHg^{-1}H=lH$, we deduce that $1=gg^{-1}\in lH$. This implies that $l\in H$ and $gHg^{-1}H=H$, we deduce that for every $h\in H$, $ghg^{-1}=ghg^{-1}1\in gHg^{-1}H=H$. This implies that $gHg^{-1}=H$ and $H$ is normal.

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  • $\begingroup$ Could you explain why $\fbox{$1\in lH$}$ ? $\endgroup$ Apr 2, 2019 at 16:21
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    $\begingroup$ @YadatiKiran it holds as $1\in H$. That is, we are assuming that $ghg^{-1}H\subset lH$ for all $h\in H$, so take $h=1$. $\endgroup$
    – user1729
    Apr 2, 2019 at 16:28
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    $\begingroup$ $gg^{-1}=g1g^{-1}1\in gHg^{-1}H$. $\endgroup$ Apr 2, 2019 at 16:30
  • $\begingroup$ Looks good, thanks! $\endgroup$
    – Gabi G
    Apr 2, 2019 at 16:37
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Aliter:

Assuming the negation of the conclusion, it follows that $(aH)(bH) = (ab)H$ (since cosets partition $G$ and $ab\in (aH)(bH)$). It follows that $H(aH) = aH$. If $h\in H$, then $hah\in H(aH) = aH$ so that $ha = ah'$ for some $h'\in H$. Hence $Ha\subseteq Ha$. Now since both left cosets and right cosets partition $G$, no inclusion can be proper. Hence, $Ha = aH$ for every $a\in G$ so that $H$ is normal.

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