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Theorem: If $\alpha:G \to H$ is a homomorphism, then $G/ \ker{(\alpha)}$ is isomorphic to $\alpha(G)$.

Denote $\ker(\alpha)$ as $K$, and recall that $K$ is a normal subgroup of group $G$ for homormorphism $\alpha$. Then the function $\beta:G/K \to \alpha(G)$ defined by $\beta(aK)=\alpha(a)$ is an isomorphism.

First, show that $\beta(aK)$ is an unambiguous/well-defined expression. Suppose the cosets $aK=bK$ for $a,b \in G$. Is $\beta(aK)=\alpha(a)=\alpha(b)=\beta(bK)$?

Recall that $g_1 \equiv g_2 \mod H \leq G$ means $g_1^{-1}g_2 \in H$, and congruence mod some subgroup $H$ is an equivalence relation on $G$, with equivalence classes being the left cosets $gH=\{gh \mid h \in H\}$ for elements $g \in G$. Similarly, defining $g_1 \equiv g_2 \mod H$ as $g_1g_2^{-1} \in H$ is also an equivalence relation partitioning $G$ into right cosets $Hg$.

The proof I'm reading (Lee's Abstract Algebra) claims this:

Suppose that $aK=bK$. Then $a^{-1}b \in K$, and therefore $\alpha(a^{-1}b)=e$. That is, $(\alpha(a))^{-1}\alpha(b)=e$, so $\alpha(a)=\alpha(b)$.

I agree that we have shown that $\alpha(b)$ is the right inverse of $(\alpha(a))^{-1}$. This is very close to $\alpha(b)=((\alpha(a))^{-1})^{-1}$, but don't we need to show $\alpha(b)$ is also the left inverse of $\alpha(a)^{-1}$?

We used the left coset equivalence relation: equivalence classes $aK=[a]=[b]=bK$, so $a \sim b$, so $a^{-1}b \in K$. We could also say $b \sim a$, so $b^{-1}a \in K$ and $\alpha(b)^{-1}\alpha(a)=e$, but I don't think this helps.

I think we have to use the fact that $K$ is normal: $Ka=Kb$, so $b \sim a$ under the right coset equivalence relation, so $ba^{-1} \in K$, so $\alpha(b)\alpha(a)^{-1}=e$, so $\alpha(b)$ is also the left inverse of $\alpha(a)^{-1}$. Am I correct, and is there a simpler way to show $\alpha(a)=\alpha(b)$?

jskattt797
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Say $g,h \in G$, and we know that $g^{-1}h = e$. Then, we have $g (g^{-1}h) = g e$. Hence, we get that $(g g^{-1}) h = h = g$

In your question, you acknowledge that $\alpha (a)^{-1} \alpha(b) = e $. You can just proceed as above from there.

In lieu of your remark about left and right inverses - those are the same in a group. If $gh = e$, then $ghg = g$, and left multiplying by $g^-1$ gives you that $hg = e$.

This is a more general fact that may interest you.

praeseo
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  • And using the exact same proof: if $hg=e$, then $hgh=h$, and left multiplying by $h^{-1}$ gives $gh=e$. So in any group, $h$ is a right inverse of $g$ iff $h$ is a left inverse of $g$. – jskattt797 May 27 '20 at 20:01
  • For $g^{-1}h=e \implies h=g$, another formulation is to see $g^{-1}h=g^{-1}g$. – jskattt797 May 27 '20 at 20:07
  • Right! There's a similar thing with the identity too, if you want - any left identity is neccasarily the identity, and likewise for right. – praeseo May 28 '20 at 04:17