## Ring of endomorphisms (2)

Posted: June 8, 2010 in Noncommutative Ring Theory Notes, Ring of Endomorphisms
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Example 2. $\text{End}_R(R^n) \cong \mathbb{M}_n(R^{op}).$

Proof. This is just an obvious result of Example 1 and Theorem 1.

Example 3. Let $G$ be a cyclic group. If $|G| = \infty,$ then $\text{End}_{\mathbb{Z}}(G) \cong \mathbb{Z}$ and if $|G|=n,$ then $\text{End}_{\mathbb{Z}}(G) \cong \mathbb{Z}/n \mathbb{Z}$.

Proof. The first part is obvious by Example 1. So suppose that $|G|=n$ and let $g$ be a generator of $G$ and $f \in \text{End}_{\mathbb{Z}}(G).$ Let $f_i(x)=x^i.$ Then, since $x^{ni}=1,$ we can choose $i$ anything we like. Now define $\varphi : \mathbb{Z} \longrightarrow \text{End}_{\mathbb{Z}}(G)$ by $\varphi(i)=f_i.$ See that $\varphi$ is an onto ring homomorphism and $\ker \varphi = n \mathbb{Z}.$ (Note that $f_0 = 0_{\text{End}_{\mathbb{Z}}(G)}.$)

Example 4. Let $G_1,G_2, G$ be cyclic groups of order $m,n, \gcd(m,n)$ respectively. Then $\text{Hom}_{\mathbb{Z}}(G_1,G_2) \cong G,$ as abelian groups.

Proof. Let $g_1,g_2$ be generators of $G_1,G_2$ respectively. Let $f_i : G_1 \longrightarrow G_2$ be defined by $f(g_1)=g_2^i.$ See that $f_i \in \text{Hom}_{\mathbb{Z}}(G_1,G_2)$ if and only if $n \mid mi$ which is equivalent to $\frac{n}{\gcd(m,n)} \mid i.$ So there are $\gcd(m,n)$ possibility for $f_i.$ Let $g$ be a generator of $G$ and define  $\varphi : G \longrightarrow \text{Hom}_{\mathbb{Z}}(G_1,G_2)$ by $\varphi(g)=f_1$ and see that $\varphi$ is a group isomorphism.

Example 5. Let $p$ be a prime number and $G_1$ and $G_2$ be cyclic groups of orders $p$ and $p^2$ respectively. Then $|\text{End}_{\mathbb{Z}}(G_1 \times G_2)| = p^5.$

Proof. By Theorem 1:

$\text{End}_{\mathbb{Z}}(G_1 \times G_2) \cong \begin{pmatrix} \text{End}_{\mathbb{Z}}(G_1) & \text{Hom}_{\mathbb{Z}}(G_2,G_1) \\ \text{Hom}_{\mathbb{Z}}(G_1,G_2) & \text{End}_{\mathbb{Z}}(G_2) \end{pmatrix}.$

By Examplse 3, $|\text{End}_{\mathbb{Z}}(G_1)|=|G_1|=p$ and $|\text{End}_{\mathbb{Z}}(G_2)|=|G_2|=p^2.$ Also, by Example 4

$|\text{Hom}_{\mathbb{Z}}(G_1,G_2|=|\text{Hom}_{\mathbb{Z}}(G_2,G_1)|=p.$

So $|\text{End}_{\mathbb{Z}}(G_1 \times G_2)|=p^5.$