So I'm going through example 15.10 in Fraleigh, which is computing $G/H$, where $G = \mathbb Z_4\times\mathbb Z_6$ and $H = \langle (0,2) \rangle$.

We have $H =\{(0,2), (0,4), (0,0)\}$, so the subgroup generated by $H$ has order $3$.

Since $G$ has $24$ elements and each coset has the same number of elements as $H$, there are $24 / 3 = 8$ cosets in $G/H$.

Fraleigh says,
The first factor $\mathbb Z_4$ of $G$ is left alone. The $\mathbb Z_6$ factor is essentially collapsed by a subgroup of order $3$, giving a factor group in the second factor of order $2$ that must be isomorphic to $\mathbb Z_2$. So $G/H$ is isomorphic to $\mathbb Z_4\times\mathbb Z_2$.

The bolded is what confuses me. Here are the elements of our factor group (the cosets of $H$ in $G$):

$(0,0) + H = H$

$(0,1) + H = \{(0,1), (0,3), (0,5)\}$

$(1,0) + H = \{(1,0), (1,2), (1,4)\}$

$(1,1) + H = \{(1,1), (1,3), (1,5)\}$

The process continues and we have $4$ more cosets.

I don't see how Fraleigh computes this factor group so quickly / without writing out the cosets. Even with writing out the cosets, I'm not sure why it's clear that $G / H$ is isomorphic to $\mathbb Z_4\times\mathbb Z_2$.

Any help much appreciated, Mariogs


marked as duplicate by Dietrich Burde abstract-algebra Jun 22 at 11:28

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Note: direct product of quotients is isomorphic to quotient of direct products.

More clearly: $(G_1\times G_2)/(H_1\times H_2) \cong G_1/H_1\times G_2/H_2$ where $H_i\leq G_i$.

From that point; $$\mathbb Z_4\times \mathbb Z_6/\langle(0,2)\rangle\cong \mathbb Z_4/\langle(0)\rangle\times \mathbb Z_6/\langle(2)\rangle\cong \mathbb Z_4 \times \mathbb Z_2.$$

  • $\begingroup$ Thanks for the help. Do you mind explaining the second half of your isomorphism claim (the one that ends in being isomorphic to $Z_4$ x $Z_2$? $\endgroup$ – bclayman Mar 30 '14 at 18:48
  • $\begingroup$ @Mariogs: since $<(0)>$ generates trivial group, $Z_4/(<0>)\cong Z_4$ and $<(2)>$ generates a subgroup with order $3$. So, quotient is a group with two elements which must be isomorpfic to $Z_2$. $\endgroup$ – mesel Mar 30 '14 at 18:55
  • $\begingroup$ I'm looking in Fraleigh and it looks like the method described above doesn't work for the case of $Z_4$ x $Z_6$ / $\langle (2,3) \rangle$ as this factor group is isomorphic to $Z_4$ x $Z_3$...thoughts? $\endgroup$ – bclayman Mar 31 '14 at 0:05
  • $\begingroup$ @Mariogs: $<(a,b)>\cong <a>\times <b>$ when $a$ and $b$ has relativly prime order. since $2$ and $3$ has same oreder; we can not write $<2,3>\cong <2>\times <3>$ so we can not use above argument. Sorry for not clarifying this point. $\endgroup$ – mesel Mar 31 '14 at 6:54
  • $\begingroup$ @Mariogs: I will try to write more general fact about quotient. $\endgroup$ – mesel Mar 31 '14 at 6:58

Fact $1$: If $\gcd(|a|,|b|)=1$ then $(a,b)\cong (a)\times (b)$. In that case, you can directly say $(\mathbb Z_m\times \mathbb Z_n)/<(a,b)>\cong (\mathbb Z_m/<(a)>)\times(\mathbb Z_n/<(b)>)$ (it follows from above argument).

Fact $2$: $|(a,b)|={|a||b|\over \gcd(|a|,|b|)}$.

Fact $3$: $\mathbb Z_{mn}\cong \mathbb Z_m\times \mathbb Z_n$ if and only if $\gcd(m,n)=1$.

Now let's try find $(\mathbb Z_4\times \mathbb Z_6)/<(2,3)>$. Notice that we can not use Fact $1$ since $2$ and $3$ have order $2$ in $\mathbb Z_4$ and $\mathbb Z_6$ respectivly.

We can say that $\mathbb Z_4\times \mathbb Z_6=<(1,0),(0,1)>$ and notice it is spanned by elements of $(1,0)$, $(0,1)$ with integer coefficients (and of course evaluated mod $(4)$ and mod $(6)$). Can we find another basis ?

You can easily see that $<(2,3),(1,1)>=\mathbb Z_4\times \mathbb Z_6$ since $\begin{bmatrix} 2 & 3 \\1 & 1 \end{bmatrix}$ can be row reduced to identity matrix with integer coefficients.

$<(2,3),(1,1)>/<(2,3)>\cong <(1,1)>$ so it is a cyclic group and by Fact $2$ $|(1,1)|={6\cdot4\over2}=12$. So, $<(1,1)>\cong \mathbb Z_{12}\cong \mathbb Z_3\times \mathbb Z_4$.

Let have one more example.

$(\mathbb Z_4\times \mathbb Z_8)/<(2,4)>$. Again we can write $(\mathbb Z_4\times \mathbb Z_8)=<(1,2),(1,1)>$ and then $<(1,2),(1,1)>/<(2,4)>\cong \mathbb Z_2\times <(1,1)>$ and $|(1,1)|={8\cdot4\over 4}=8$ so

$$(\mathbb Z_4\times \mathbb Z_8)/<(2,4)>\cong \mathbb Z_2\times \mathbb Z_8.$$


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