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?

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

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  • $\begingroup$ I actually knew that one. I guess it was the crazy pills. Thank you! EDIT: So if it was Z11 x Z15, then (1,1) would be a generator, and the factor group (Z11 x Z15)/(<(1,1)>) would be order 1.... ? $\endgroup$ – Indigo Dec 2 '15 at 9:47
  • $\begingroup$ @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. $\endgroup$ – Ben Grossmann Dec 2 '15 at 10:02
  • $\begingroup$ 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? $\endgroup$ – Indigo Dec 2 '15 at 10:14
  • $\begingroup$ @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 $p$. $\endgroup$ – Ben Grossmann Dec 2 '15 at 10:18
  • $\begingroup$ I don't understand why $0+\langle 4 \rangle$ is order 1, if $\langle 4 \rangle$ itself is order 3. $\endgroup$ – 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"

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