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Suppose you have $G$ a group, and $H,K$ both subgroups of $G$. Let $f: H \times K \rightarrow G$ such that $f((h,k))=hk$. Show that $H \times K /\sim_f$ is in bijection with $H \cap K$. Where $\forall (x_1,x_2) \in E^2, (x_1 \sim_f x_2 \iff f(x_1) = f(x_2))$

I have a hard time doing this. I saw there is a similar question but I am unable to comprehend it. My teacher gave me a proof, by trying to find an element such that if $hk = h'k'$ then by putting $t=h^{-1}k'$ you get $h'k' = (hl)(l^{-1}k)=h(ll^{-1})k=hk$, but I can't understand how this proves the bijection.

John Mayne
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  • Are you sure the problem says $H\cap K$, not $HK$. The latter would make the statement correct, but as it stands, with $H\cap K$, it's not true in general. A small counterexample is $G=\mathbb Z/2\times\mathbb Z/2$, $H=\mathbb Z/2\times{0}$, and $K={0}\times\mathbb Z/2$. – Andreas Blass Sep 23 '17 at 16:59
  • I don't think this is true. What if $G$ itself is the (internal) direct product of $H$ and $K$? Then $f$ is a bijection between $H \times K$ and $G$, and $H \cap K = 1$. –  Sep 23 '17 at 16:59

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First thing you should do is verify that $\sim_f$ is an equivalence relation. Remember that the elements in $H\times K/\sim_f$ are classes $[h,k]$.

Next, try to find the function that determine the bijection between $H\times K/\sim_f$ and $H\cap K$. Looks like your function is $g([h,k])=f(h,k)=hk$. Everytime you define a function over a quotient you must guarantee is well defined by taking two elements of the same class: $[h,k]=[h^\prime,k^\prime]$ implies $g([h,k])=g([h^\prime,k^\prime])$. This is the case because the relation was defined like that. Injectivity is easy to see using the same arguments in the opposite direction. For surjectivity, every element $x\in H\cap K$ is the image of $[e,x]$ (or $[x,e]$ both of them are on the same class), where $e$ is the identity of the group.

Juan C. Cala
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