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

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?

• From the fact that 1 generates Z2 and Z4 you can only conclude that $\{(1,0),(0,1)\}$ generates the product. – Justpassingby Dec 2 '15 at 9:39
• The product of cyclic groups is not necessarily cyclic. – Arthur Dec 2 '15 at 9:40
• Pretty sure the order is $4$. The element $(1,2)$ can not be generated by your $(1,1)$ – IAmNoOne Dec 2 '15 at 9:42
• you did it right but note that while the second 1 is trying to generate the second group the first group was done. – mrs Dec 2 '15 at 9:43
• So it seems I was just confused about what is required for an element to generate the cross product. Thanks everyone. – Indigo Dec 2 '15 at 9:46

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.
• @Indigo yes, that's right. In particular, since $\langle (1,1) \rangle = G$, your factor group is $G/G$, which is just the trivial group. – Ben Grossmann Dec 2 '15 at 10:02
• So, then, say we had $\mathbb{Z}_{12} / \langle 4 \rangle$, and wanted the order of an element $5+\langle 4\rangle$? Since $\langle 4\rangle$ is order 3, wouldn't any element of the factor group also be order 3? – Indigo Dec 2 '15 at 10:14
• @Indigo not any element, surely $0 + \langle 4 \rangle$ has order $1$. But yes, the non-identity elements of a (cyclic) group of prime order $p$ have order $p$. – Ben Grossmann Dec 2 '15 at 10:18
• I don't understand why $0+\langle 4 \rangle$ is order 1, if $\langle 4 \rangle$ itself is order 3. – Indigo Dec 2 '15 at 10:20
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"