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This may sound like an easy question, but I never learned cyclic groups due to my time constraint of my class and that the professor ran out of time to teach it. However, he left me with a problem to solve. The question is "Show that both $2$ and $3$ are generators of the additive group $Z/(5)$. I know that a cyclic group is a group that has one element that can generate other elements of the same form in that group. But how do I show with $2$ and $3$. I thought about showing $(2+3)^n$ would make that the generator of $Z/(5)$. Is that correct? If not, can someone explain to me how it works. Thanks.

Henry Lee
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2 Answers2

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Since

$$\Bbb Z=\langle 2,3\rangle\implies \Bbb Z/5\Bbb Z=\langle 2+5\Bbb Z,\,3+5\Bbb Z\rangle$$

Everything, of course, additive. Of course, you don't need two generators of either group, but you can use two generators.

DonAntonio
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  • Would it be the same if I did $⟨25Z,35Z⟩$ if it was the multiplicative group of the field $Z/(5)$? – Henry Lee May 15 '16 at 20:20
  • @HenryLee No as $;2\cdot5\Bbb Z;$ is not even an element of $;\Bbb Z/5\Bbb Z;$ ... – DonAntonio May 15 '16 at 20:23
  • How would I do it for the multiplicative group? – Henry Lee May 15 '16 at 20:48
  • @HenryLee It's just the same exactly: $$\left(\Bbb Z_5\right)^*=\langle 2+5\Bbb Z,,3+5\Bbb Z\rangle=\langle 2+5\Bbb Z\rangle$$ If you want two generators choose first option, if you want only one choose the second one – DonAntonio May 15 '16 at 21:38
  • I'm asking for the multiplicative group of $Z/5$. – Henry Lee May 15 '16 at 21:54
  • @HenryLee That's exactly what I answered in my past comment... – DonAntonio May 15 '16 at 22:00
  • $⟨2+5Z⟩$ does not generate the entire field of $Z/(5)$ – Henry Lee May 15 '16 at 23:04
  • @HenryLee True, it doesn't. It just generates the multiplicative group $;\Bbb Z_5^*;$ ,as you asked in your second comment below my answer. – DonAntonio May 15 '16 at 23:06
  • Then I'm confused, What is a multiplicative group of a $Z_5$. Sorry, I'm new with group theory. – Henry Lee May 15 '16 at 23:12
  • @HenryLee Is the set of all elements of the field different from zero. When this set is finite we get a cyclic group, as in this case. But we also have multiplicative groups with the more well known fields: $;\Bbb Q^,,,,\Bbb R^;,;;\Bbb C^;$ and etc., though these last ones are not* cyclic (but any finite subgroup of them is) – DonAntonio May 15 '16 at 23:14
  • I need to restate the question. The question says "Let {$1,2,3,4$} be the multiplicative group of the field $Z/(5)$. Show that this group is cyclic by showing that $2$ is a generator of the group. Show that $3$ also generates the group. – Henry Lee May 15 '16 at 23:16
  • @HenryLee That's the question I understood all along and the one I addressed and answered...Take into account that in this case $;{1,2,3,4};$ is not a subset of the natural numbers but in fact a set of representatives of equivalence clases modulo $;5;$ different from the equiv. class corresponding to zero. Sometimes these are written as $;{\overline1,,\overline2,...};$ . As already mentioned, this group is cyclic since the field is finite, and since (doing arithmetic modulo five): $;2,2^2=4=-1,,2^3=8=3,,2^4=1;$, we see $;\langle\overline2\rangle;$ indeed generates . – DonAntonio May 16 '16 at 07:43
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Simplest proof: $2, 2+2=4, 2+2+2=1, 2+2+2+2=3, 2+2+2+2+2=0$. Hence $2$ generates the additive group $Z/(5)$, which has 5 elements. The proof that $3$ generates it is similar.

vadim123
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