# General Algorithm for Computing Factor Groups

I'm a bit confused on computing factor groups. Fraleigh defines it as classifying a factor group according to the fundamental theorem (saying what the factor group is isomorphic to).

For instance, in Example 15.7 he has:

Compute the factor group $(Z_4$ x $Z_6)$ / $<(0,1)>$.

He tells us that:

H = ${(0,0), (0,1), (0,2), (0,3), (0,4), (0,5)}$ which makes sense to me.

$(Z_4$ x $Z_6)$ has 24 elements and $H$ has 6, so all cosets of $H$ must have 6 elements and $(Z_4$ x $Z_6)$ / $H$ must have order 4.

Since $(Z_4$ x $Z_6)$ is abelian, so is $(Z_4$ x $Z_6)$ / $H$

He lists out the cosets:

${(0,0) + H, (1,0) + H, (2,0) + H, (3,0) + H}$

Then he claims that it's clear that our factor group is isomorphic to $Z_4$...

1) Why is this the case?

2) Is there a general way of going about computing factor groups? It's not totally clear to me how Fraleigh does this.

Thanks for the help, Mariogs

($H$ being a subgroup guarantees that 'being in the same (right) coset' is an equivalence relation: i.e. it guarantees $a=b=c\implies a=c\,\land\,b=a$ in the quotient group, and $H$ being normal subgroup implies that $\ a=b,\ c=d\ \implies\ ac=bd\$ and $\ a=b\implies a^{-1}=b^{-1}$).
Now, in the example, all elements $(a,x)\in\Bbb Z_4\times\Bbb Z_6$ are in the same coset with any other $(a,y)$. So, basically the coset $(a,0)+H$ correspond to $a\in\Bbb Z_4$.
This $(a,x)\mapsto a$ gives a concrete isomorphism from the quotient group to $\Bbb Z_4$, as we have
A homomorphism $f:G\to G'$ factors through the quotient group $\,G/N\$ iff $\ N\subseteq \ker f\$ [i.e. $f(n)=e$ for each $n\in N$ where $e$ is the unit of $G'$].
(Since $f$ is a homomorphism, it means exactly that equal elements in the quotient are mapped to equal elements.)
• I agree that there are four elements in our factor group, but why does this mean it's isomorphic to $Z_4$? I suppose in this case it's clear that it's not isomorphic to the Klein 4-group since there are elements with order > 2... – bclayman Mar 30 '14 at 15:54