I am actually doing a problem in hatcher, and when I check the solution here: http://www.math.wisc.edu/~ccheng/courses/Hatcher.pdf
I find a line says (on the bottom line of page 39):
$\Bbb Z^n/ \ker (f) = \Bbb Z$, so $\ker (f) = \Bbb Z^{n-1}$
It seems quite natural and intuitive, but we are never taught that we can manipulate quotient groups just in the same manner of quotient of numbers and I am stuck on proving it. So may I please ask if the statement "if $A/B=C$, then $B=A/C$ for groups $A,B,C$" holds in general? If it is true, could someone give a proof? If it is not true, may I please ask what is special here to make this sentence hold?
Thanks!
Edit: I have find a counter-example $\Bbb Z/n\Bbb Z=\Bbb Z/n\Bbb Z$ but $\Bbb Z/n\Bbb Z$ is not contained as a copy in $\Bbb Z$ so we cannot quotient it out. But I have see something quite like doing manipulation of groups in this manner recently and now I am confused. So may I please ask under what condition could we do so?
First of all, let me point our that technically speaking the statement "$\mathbb{Z}^n/\ker(f)=\mathbb{Z}$" is not true, or rather is a little bit inaccurate. A more accurate statement would be that "$\mathbb{Z}^n/\ker(f)\cong\mathbb{Z}$". But in some contexts it's a common practice and an acceptable abuse of notation to identify isomorphic groups, so let's turn a blind eye to that.
To answer your question: no, it doesn't hold in general, simply because $C$ may not be a subgroup at all, even up to isomorphism. Quick example: if $A=\mathbb{Z}$ and $B=2\mathbb{Z}$, then $C=A/B=\mathbb{Z}/2\mathbb{Z}=\mathbb{Z}_2$; but we can't form $A/C$, because in $A=\mathbb{Z}$ there's no subgroup isomorphic to $C=\mathbb{Z}_2$.
It does work, though, for abelian groups when $A$ splits into a direct sum, i.e. when $A=B\oplus C$; then both $A/B=C$ and $A/C=B$ (again, with a slight abuse of notation). The same goes for nonabelian groups, but we'd speak of direct products then.