Compute the quotient group $(\mathbb{Z_{4}} \times \mathbb{Z_{8}})/ \langle (3,3) \rangle$.

Question

Compute the quotient group $(\mathbb{Z_{4}} \times \mathbb{Z_{8}})/\langle (3,3)\rangle$.

Let $H=\langle(3,3)\rangle$. We find that $|H|=8$. Then, the order of the factor group is $|(\mathbb{Z_{4}} \times \mathbb{Z_{8}})/H| \implies \frac{4 \cdot 8}{8} =4$. We find that the abelian groups are: $\mathbb{Z_{4}}, \mathbb{Z_{2}} \times \mathbb{Z_{2}}.$

Note that the coset $(1,0) + H$ has order $4$ in $(\mathbb{Z_{4}} \times \mathbb{Z_{8}})/H$, i.e. we must find the smallest positive $n$ such that $n(1,0) \in H$. Afer attempting all elements in the set, we find that the only element to satisfy this condition is $(\bar{n},0)=(0,0)$, i.e.

$\bar{n} \equiv \bar{0} \mod 4$ $\implies$ $n=4$. Hence, $|(1,0) + H|=4$.

Thus, $(\mathbb{Z_{4}} \times \mathbb{Z_{8}})/H\simeq \mathbb{Z_{4}}$.

The problem arises when attempting to show that the coset $(1,1)+ H$ is a generator of $(\mathbb{Z_{4}} \times \mathbb{Z_{8}})/H$. We must find the order of $(1,1)+ H$: the smallest positive $n$ such that $n(1,1) \in H$. Again, the only possible choice which should get us an order of $4$ is $(\bar{n},\bar{n})=(0,0)$, then

if $4|n$ and $8|n$ $\implies$ $n=\text{lcm}(4,8)=8$

This is where the problem arises as we want $n=4$ not $8$. Any suggestion and help would be greatly appreciated.

Answer 1

Note that in $G$ we have $(3,3)+(3,3)+(3,3)=(1,1)$ and thus it is in $H$ so it's in the kernel under the canonical projection and therefore cannot generate $G/H$.

Answer 2

The order of $(1,1)+H=H$ is $1$.

The (two) generators of $\Bbb Z_4$ in this case are $(1,0)+H$ and $(0,1)+H$.