# Is this true for cosets of normal subgroups?

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.

• 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. Feb 23, 2020 at 17:30
• 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 $Hb=bH = aH = b^{-1}H=Hb^{-1}$... Feb 24, 2020 at 2:45