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If $N$ is the least normal subgroup of $A*B$ containing $A$, then $(A*B)/N \cong B$.

My Proof: Let $f:A \to B$ be the homomorphism given by $f(a) = e_B$. Note that $1_B:B \to B$ is also a homomorphism. Therefore, by the Universal Property of $A*B$, there exists a unique group homomorphism $\eta:A*B \to B$ where $\eta \circ i_A = f$ and $\eta \circ i_B = 1_B$ (where each $i_X$ is the inclusion map $i_X:X \to A*B$). Since $1_B = \eta \circ i_B$ is surjective, so is $\eta$. Additionally, note that $\ker(\eta) = \langle \langle A \rangle \rangle = N$ (I have no idea whether this is true or not). Therefore, by the First Isomorphism Theorem of Group Theory, there exists an isomorphism $(A * B)/\ker(\eta) \to B$. Therefore, $(A * B)/N \cong B$.

The part of the proof I'm having trouble with is finding a suitable homomorphism $f:A \to B$ so that $\ker(\eta) = N$. In particular, I am unsure how to show that $\ker(\eta) = N$ in any case. There is a question here asking about what the least normal subgroup is; however, it never gained any traction. Additionally, there is another question If $N$ is the normal subgroup of $A\ast B$ generated by $A$, then $(A\ast B)/N\cong B$, which is similar to mine.

Is my proof on the right track? How do I find a suitable homomorphism $f$ and how would I prove that $\ker(\eta) = N$? Thanks for any help.

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  • $\begingroup$ In “Therefore” just before “The part of the proof”, I think you mean “$(A*B)/\mathrm{ker}(\eta)$” $\endgroup$ Apr 1, 2020 at 17:59
  • $\begingroup$ Yep, thank you @ArturoMagidin. $\endgroup$
    – HiMatt
    Apr 1, 2020 at 18:01
  • $\begingroup$ You don’t know that $\mathrm{ker}(\eta)=N$... yet. $\endgroup$ Apr 1, 2020 at 18:01
  • $\begingroup$ I don't know either (hence, why I am here). I denote the least normal subgroup as $\langle\langle A \rangle\rangle = N$. $\endgroup$
    – HiMatt
    Apr 1, 2020 at 18:03
  • $\begingroup$ Yet your proof says “note that $\mathrm{ker}(\eta)=\langle\langle A\rangle\rangle =N$”. So... do you know the kernel is the normal closure of $A$, or don’t you? $\endgroup$ Apr 1, 2020 at 18:04

1 Answer 1

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Your kernel contains $A$, and therefore must contain $N$. Hence there is a surjective map $A*B/N \to B$. Moreover, $BN = A*B$ (as the lhs is a subgroup and contains $A$ and $B$), and by the second isomorphism theorem, $A*B/N = B/(N\cap B)$. Composing with our first map, we obtain a surjective map $B/(N\cap B) \to B$ which factors through $A*B/N$. But looking at the definitions of our maps, this map is just $b \mapsto b$, whence $N\cap B\cong \langle e\rangle$ and everything is an isomorphism. In particular $A*B/N \cong B$

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  • $\begingroup$ I have a few questions. The detail $BN = A * B$ is not so obvious to me. You don't mean $B*N = A*B$ by chance (which wouldn't make sense but that is all I can think of)? And how does looking at the definitions of our maps show that $b \mapsto b$? If we have $B/(N \cap B) \to (A*B)/N \to B$, then are you saying $(N \cap B)b \mapsto b$ for all $b \in B$ and hence, $(N \cap B) = \langle e \rangle$? Thanks for the help! $\endgroup$
    – HiMatt
    Apr 1, 2020 at 19:54
  • $\begingroup$ No I really mean $BN = A*B$. Which part of the claim doesn't make sense? And yes I mean $b(N\cap B)\mapsto b$ $\endgroup$
    – vujazzman
    Apr 2, 2020 at 0:18

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