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I have this question, where, we are given that $H$ is a subgroup of $G$, and that $|G : H| = 2$. Now we suppose that $a, b \in G$ but $a, b \notin H$. Why is it true that $ab \in H$?

I know that $|G : H| = 2$ implies the number of distinct left cosets is 2. I think I'm supposed to use lagrange's theorem but I don't know how to apply it here. Any idea why the claim is true?

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Isn't $H = 2\mathbb{Z}$ a subgroup of $G=\mathbb{Z}$? Then let $a=1, b=3$


in general $cH = H$ or $cH \cap H = \varnothing$. In the second case, so that $G=H \cup cH$. Then if $a,b \in cH$ then $ab \in cH cH$.

Then $\big(cH c^{-1} \big) \big( c^{-1} cH \big)= H \, H = H$

this proof lacks an explicit nature in my case, in particular check that $cH = H c $ but this is definition of subgroup

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  • $\begingroup$ Is it (Z,+) here $\endgroup$ – Tsemo Aristide Nov 7 '16 at 14:58
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If $a,b\notin H$ then $b\in a^{-1}H$ (clear because $[G:H]=2$). Then $b = a^{-1}h$ for some $h\in H$. Then $ab = h$, so $ab\in H$.

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$H$ is normal, Let $p:H\rightarrow G/H$, so $a,b$ not in $H$ implies that $p(a)=p(b)$ is the generator of $Z/2$ so $p(a)p(b)=1=p(ab)$ and $ab\in H$.

https://proofwiki.org/wiki/Subgroup_of_Index_2_is_Normal

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Like you said, there are only two cosets. If $a, b \notin H$ then $b = ah$ for some $h \in H$, so if their product $ab = a^2h \notin H$ we have $a^2h = ah'$ for some $h' \in H$. Simplifying gives $a h = h'$ so $a \in H$, a contradiction.

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