This is about correct operation in a multiplicative cyclic group. Say we have the group $\mathbb Z_4^{\times} = \{1,2,3\}$, if we multiply the elements, say $3\cdot 2\cdot 2 $ we get $12 \mod 4 = 0$ but $0\notin \mathbb Z_4^{\times}$ so that can't be right, so what is the correct result of this operation?

  • $\begingroup$ Notice that $(4,2) = 2 \neq 1$, thus $2 \notin \mathbb Z^{\times}_4$. $\endgroup$ – Aaron Maroja Feb 5 '16 at 14:42

You're using $^\times$ in correctly. $S^\times$ does not mean "remove zero". It means "remove all non units". In some contexts this is the same, because your set is a field, where every nonzero element is invertible.

Definition: $x\in R$ is a unit (of $R$) if $xy=1_R$ has a solution. Another way of saying this is to ask for the invertible elements.

Definition: Let $R$ be a ring (with identity). $R^x\subset R$ is the set of all units of $R$. It forms a group under the multiplication operation of $R$.

Examples: $$\mathbb{Z}^\times=\{-1,1\}$$ $$\mathbb{Z}_m^\times=\{a\in\mathbb{Z}_m:(a,m)=1\}$$ $$\mathbb{Z}[i]^\times=\{1,i,-1,-i\}$$

In this specific case, $2$ is not an element of the group because it is not a unit, i.e. The equation $2x=1$ has no solutions. The correct unit group is $\{1,3\}\sim C_2$.

  • $\begingroup$ What does it mean "not a unit"? isn't there only one unit element, in this case $1$ ? and what makes $3$ a "unit" ? $\endgroup$ – shinzou Feb 5 '16 at 14:40
  • $\begingroup$ @kuhaku edited with more information. $\endgroup$ – Stella Biderman Feb 5 '16 at 14:44

We have $\mathbb{Z}_4^\times = \{1,3\}$, since $2$ is not a unit in $\mathbb{Z}_4$.


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