Anonymous at Tue, 7 May 2024 02:03:15 UTC No. 16163375

>>16163364

Let $H$ be an index 2 subgroup of a group $G$, meaning there are exactly two left cosets of $H$ in $G$. Let these cosets be $H$ and $gH$, where $g \in G$ but $g \notin H$.

To show that $H$ is normal in $G$, we need to prove that for any $g \in G$, $gH = Hg$.

Since there are only two left cosets, $gH$ must be the remaining coset, which is $G \setminus H$. Similarly, $Hg$ must be the remaining coset.

Now, let's consider the element $hg$ where $h \in H$. Since $H$ is a subgroup, it is closed under the group operation. So, $hg$ must be in either $H$ or $gH$.

If $hg \in H$, then $g = h^{-1}(hg) \in H$, contradicting the assumption that $g \notin H$.

Thus, $hg$ must be in $gH$. This means that for any $h \in H$ and $g \in G$, $hg$ is in $gH$.

Therefore, $gH = Hg$, and $H$ is normal in $G$ since all left cosets are also right cosets.

Anonymous at Tue, 7 May 2024 02:08:05 UTC No. 16163382

G < H < G
H = G
Qed

Anonymous at Tue, 7 May 2024 03:05:26 UTC No. 16163422

>>16163364
Let g \in G-H so that G = H \cup gH (disjoint). If gHg^{-1} \ne H then ghg^{-1} = gh' for some h,h' \in H, but then h'^{-1}h = g and g \in H, contradiction.

>>16163375
>5 pages of rambling for a one line proof
I hate undergrads so fucking much

Anonymous at Tue, 7 May 2024 12:02:15 UTC No. 16163852

>>16163422
R.I.G.O.U.R

you were undergrad too once, buddy