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In this answer to a question about the homology of the Klein bottle, it is said that we can "simplify" the quotient

$\langle a,b,c\rangle/\langle a+b-c,a-b+c\rangle$

to

$\langle a+b-c,b,c\rangle/\langle a+b-c,2b-2c\rangle$,

where these are the free abelian groups generated by the respective elements.

I think that this means that we're saying that the two groups are isomorphic, but I have not come across this sort of simplification before.

I can see that the generators have been manipulated to become linear products of the generators which we already have (for example, $2b - 2c = (a + b - c) - (a-b+c)$). This intuitively makes some sense to me, but it's not entirely clear to me why this results in an isomorphic group.

What I'm thinking is that given such a manipulation of the generators, we can construct an explicit isomorphism between the two groups. Is that how this works? If so, how do we do that, and prove that it's an isomorphism?

Matt
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1 Answers1

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Define a map $\phi\colon \langle a,b,c\rangle \rightarrow \langle a,b,c \rangle$ of the free abelian groups, defined on the generators by $\phi(a)=a+b-c$, $\phi(b)=b$ and $\phi(c)=c$. Clearly $\phi$ is bijective. It is given on the generators, so it extends to a group isomorphism. So we have $\langle a,b,c\rangle \cong \phi(\langle a,b,c\rangle )=\langle a+b-c,b,c\rangle$. It clearly induces also an isomorphism on the quotients, because the normal subgroups are isomorphic.

Dietrich Burde
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