Lagrange's Theorem: Injectivity.

Question

Let $U$ be a subgroup of the finite group $G$ and $a\in G$.

To show Lagrange's Theorem we defined a map $U \rightarrow aU$, $u\mapsto au$ and said that it was bijective. Now for the injectivity we used the trivial case that for $au_1 = au_2\Leftrightarrow u_1=u_2 $ but this uses that we add $a^{-1}$ from the left side in both cases.

Now my question is that of course there exists a $a^{-1}\in G$ but don't we need to have $a^{-1}\in aU$ to add it from the left? And if so, how can we show $a^{-1}\in aU$?

Best, KingDingeling

Answer 1

Okay so essentially what you're trying to show is that if $au_1=au_2$ then $u_1=u_2$, and you seem to think that we are restricted completely to working within $aU$ to do this. But this isn't the case at all, in fact since $u_1$ and $u_2$ may not be in $aU$ (they are not in the case that $a\notin U$), so it doesn't even make sense that we could show $u_1=u_2$ while working entirely within the set $aU$.

The point is that these equalities we are worried about are occurring within $G$, so it is valid to use any element of $G$ to show this claim.

Answer 2

You don’t need to have $a^{-1} \in aU$ and won’t it by the way in mot cases.

This is by the way not required to prove the injectivity. The argument using $a^{-1}$ is not making that hypothesis.