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I see it claimed that $(\mathbb Z/2\mathbb Z)^*$ is a cyclic group, but it doesn't seem to me that it has a generator. It is the set $\{0,1\}$, both elements of which are their own square. So it seems to me that each element has order $1$ and the group is not cyclic. Why is this a cyclic group?

Myungheon Lee
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Daniel Johnston
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1 Answers1

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The multiplicative group of a group $G$ is the collection of all those elements of the group which have a multiplicative inverse. In your case $\Bbb{Z}/2\Bbb{Z}$ does $\bar{0}$ have an element $\bar{a} \in \Bbb{Z}/2\Bbb{Z}$ such that $\bar{0a} = 1$ ?

Note that this is true for $\Bbb{Z}/n\Bbb{Z}$ for any $n$. In general you define the multiplicative group of $G$ (or the group of units, as its sometimes called) on $G-\{0\}$

Sayan Chattopadhyay
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    What is true for any $n?$ – suckling pig Nov 20 '24 at 21:32
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    that the group of units for $\mathbb{Z}/n\mathbb{Z}$ omits $0$. – NewViewsMath Nov 20 '24 at 22:33
  • Probably worth noting that not all such groups are cyclic, since the phrasing at the end mind be confusing. – Derek Allums Apr 10 '25 at 07:02
  • The group of units does include $0$ when $n=1$. So the claim is only true for $n\ge2$. – Emil Jeřábek Apr 10 '25 at 07:14
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    More importantly, there is not such thing as “the multiplicative group of a group $G$”. A group $G$ carries only one operation, and all elements are invertible wrt that. It only makes sense to speak of the multiplicative group of a ring $R$. And the domain of the multiplicative group in that case is not $R\smallsetminus{0}$, but ${x\in R:\exists y\in R,xy=yx=1}$, which, on the one hand, will omit elements other than $0$ if $R$ is not a field (or division ring), and on the other hand, will include $0$ if $R$ is the trivial ring. – Emil Jeřábek Apr 10 '25 at 07:20