Exercise 4.4.16 ( $(\mathbb{Z}/24\mathbb{Z})^ imes$ is an elementary abelian group of order $8$)

Question

Prove that $(\mathbb{Z}/24\mathbb{Z})^	imes$ is an elementary abelian group of order $8$. (We shall see later that $(\mathbb{Z}/n\mathbb{Z})^	imes$ is an abelian elementary abelian group if and only if $n\mid 24$.)

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

\begin{proof} Here we write $k$ for the class $[k]_{24}$ of $k$ modulo $24$. Then
$$(\Z/24\Z)^	imes = \{1,5,7,11,-1,-5,-7,-11\}$$
has order $\varphi(24) = 8$.

Since $$1^2 \equiv 5^2  \equiv 7^2 \equiv 11^2 \equiv (-1)^2 \equiv  (-5)^2  \equiv (-7)^2 \equiv  (-11)^2 \equiv 1 \pmod {24},$$ every element of the group $(\Z/24\Z)^	imes$ has order $1$ or $2$, thus $(\mathbb{Z}/24\mathbb{Z})^	imes$ is an elementary abelian group of order $8$, isomorphic to the additive group $(\Z/2\Z)^3$.

\end{proof}

Exercise 4.4.16 ( $(\mathbb{Z}/24\mathbb{Z})^\times$ is an elementary abelian group of order $8$)

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Exercise 4.4.16 ( $(\mathbb{Z}/24\mathbb{Z})^\times$ is an elementary abelian group of order $8$)

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