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I was wondering under what conditions cosets are closed under inverses. For normal subgroups, I came up with this result:

Let $G$ be a group and $H$ be a normal subgroup of $G$. Then each coset of $H$ either contains inverses for all elements, or none at all.

More precisely, $a^{-1} \in Hb \iff b^{-1} \in Hb$. Since $b$ can be replaced with any representative of the coset, the existence of an inverse of a general element $a$ is necessary and sufficient for the existence of an inverse of any other element.

Proof: $a \in Hb$ for some $b \in G \iff a = hb$ for some $h \in H$. Then $a^{-1} = (hb)^{-1} = b^{-1}h^{-1}$, so $a^{-1} \in Hb \iff b^{-1}h^{-1} \in Hb = bH$ (since $H$ is normal) $\iff b^{-1 }h^{-1} = bh’$ for some $h’ \in H$ $\iff b^{-1} = bh’h = bh’’$ for some $h’’ \in H$ $\iff b^{-1} \in bH = Hb$.

Is this true? Is it useful? Could someone provide some examples of groups to help verify this, or otherwise provide a counterexample? I’m quite new to group theory.

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    $\begingroup$ Another way to state the condition is that the coset has order $2$ (or $1$) in the quotient for some choice of representative, but this property is of course independent of choice of representative. $\endgroup$ Feb 23, 2020 at 17:30
  • $\begingroup$ So, you are trying to prove that if $a\in Hb$, then $a^{-1}\in Hb$ if and only if $b^{-1}\in Hb$? Doesn’t that follow by noting that $x\in Hb$ if and only if $Hb=Hx$, and $Hx=Hy\iff x^{-1}H = y^{-1}H$; and so we have $Hb=Ha=Ha^{-1}$ iff $Hb=bH = aH = b^{-1}H=Hb^{-1}$... $\endgroup$ Feb 24, 2020 at 2:45

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