Find the order of $\langle (1,1)\rangle$ in $\mathbb{Z}_2 \times \mathbb{Z}_4$.

Question

I feel like I'm taking crazy pills. 1 generates $\mathbb{Z}_2$, and likewise generates $\mathbb{Z}_4$. So shouldn't (1,1) generate the whole thing? Yet I keep running up against

$\langle (1,1) \rangle = \{ (1,1), (0,2), (1,3), (0,0) \} $

which is order 4. $\mathbb{Z}_2 \times \mathbb{Z}_4$ is clearly order 8.

I looked at other examples on here to see if I was generating $\langle (1,1) \rangle$ correctly, and I THINK I am.... Am I not?

Answer 1

You are indeed writing $\langle (1,1) \rangle$ correctly, and no: $(1,1)$ does not (on its own) generate $\Bbb Z_2 \times \Bbb Z_4$.

In fact, $\Bbb Z_2 \times \Bbb Z_4$ is not generated by any single element, which is to say it is not cyclic. In general, $\Bbb Z_m\times \Bbb Z_n$ will be cyclic if and only if $m$ and $n$ are relatively prime.

Answer 2

Further to another answer, it is easy to convince yourself that: " The order of element $(a,b)\in Z_{n}\times Z_{m}$ is the l.c.m of the orders of $a$ and $b$ in their groups respectively"