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All you need to show, is that it also has a right identity, and a right inverse.

yea,,, I have hourssss trying to :(

That's true as well for non-Abelian groups, right?

Wait, yes it is. I was derping; I forgot the exact definition of Abelian for a second.

Man, I'm too sleepy for this. :(

bad reference... please... i have 3 nights with no sleep :(

:(

That took some time to take apart and figure out exactly what was going on.

im still trying to understand the inverses

Any particular step in that proof?

well,, im trying to understand step by step... let me see if i get stuck

Woops, I tried proving it backwards using the right inverse. I feel dumb.

waittt

if x is the inverse of a and if we have left inverse the will be (x*a) nooo (a*x)

If that was unclear, it means that \(x*a=e\).

And we want to show that \(a*x=e\) as well.

yea... but u are using the right one

yea but you can apply that statement like its true.. u want to get that statement

but my understanding is that u cant just say x*a=e=a*x.. u have to get the answer

ok let em check again

ok thanks you very much
I kind of getting it..