Ring of endomorphisms (2)

Posted: June 8, 2010 in Noncommutative Ring Theory Notes, Ring of Endomorphisms
Tags: , ,

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


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


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