Proving Normality Subgroups in a Group | G/H = 2 | Step-by-Step Guide

[mathusers]

Hi next one? Any ideas here?

Let G be a group and H \subset G a subgroup such that |G/H| = 2. Show that H is normal in G.

thnx :)

 

[Dick]

Did you try anything? Any ideas? Try thinking about right and left cosets.

 

[mathusers]

[G/H]=2 means that H has two left and right cosets in G. Assume g_1 \in H, then it is trivial that g_1H = H = Hg_1. Now assume g_2 \notin H, this means that g_2H \neq H \neq Hg_2. But since there are only 2 cosets and both of them are not in H then it means they are the same so g_2H = Hg_2

is this correct?

 

[Dick]

[G/H]=2 means that H has two left and right cosets in G. Assume g_1 \in H, then it is trivial that g_1H = H = Hg_1. Now assume g_2 \notin H, this means that g_2H \neq H \neq Hg_2. But since there are only 2 cosets and both of them are not in H then it means they are the same so g_2H = Hg_2

is this correct?

Yep, that's pretty much it.

 

[mathusers]

just a question here though, when it says g_1H = H = Hg_1, is it referring to g_1 as the identity element of the group? if so, can you please explain why this proves that H is normal in G?
i understand "how" to do the question but i don't understand "why" it works... thanks :)

 

[Dick]

There are two left cosets g1H and g2H. One of them is eH=H. If g1H=H, then g1 could be e, it could also be anything else in H. As you said. g2 can be anything not in H. If you really want to understand it prove these statements.