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


Forming a quotient group can be regarded as modifying the equality of the original group, such that all elements that are in the same coset will be regarded equal to each other in the quotient group, and that different cosets correspond to different elements.
($H$ 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.)

  • 1
    $\begingroup$ 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... $\endgroup$ – anon_swe Mar 30 '14 at 15:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.