Math4302 Modern Algebra (Lecture 9)

Groups

Non-cyclic groups

Dihedral groups

The dihedral group $D_n$ is the group of symmetries of a regular $n$ -gon.

(Permutation that sends adjacent vertices to adjacent vertices)

$D_n<S_n$

$|S_n|=n!, |D_n|=2n$

We can classify dihedral groups as follows:

$\rho \in D_n$ as the rotation of a regular $n$ -gon by $\frac{2\pi}{n}$ .

$\phi\in D_n$ as a reflection of a regular $n$ -gon with respect to $x$ -axis.

We can enumerate the elements of $D_n$ as follows:

D_n=\langle \phi,\rho\rangle=\{e,\rho,\rho^2,\cdots,\rho^{n-1},\phi,\phi\rho,\phi\rho^2,\cdots,\phi\rho^{n-1}\}

We claim these elements are all distinct.

Proof

Consider the first half, clearly $\rho_i\neq \rho_j$ if $0\leq i<j\leq n-1$ .

Also $\phi\rho_i\neq \phi\rho_j$ if $0\leq i<j\leq n-1$ . otherwise $\rho_i=\rho_j$

Also $\rho^i\neq \rho^j\phi$ where $0\leq i,j\leq n-1$ .

Otherwise $\rho^{i-j}=\phi$ , but reflection (with some point fixed) cannot be any rotation (no points are fixed).

In $D_n$ , $\phi\rho=\rho^{n-1}\phi$ , more generally, $\phi\rho^i=\rho^{n-i}\phi$ for any $i\in\mathbb{Z}$ .

Group homomorphism

Definition for group homomorphism

Let $G,G'$ be groups.

$\phi:G\to G'$ is called a group homomorphism if $\phi(g_1g_2)=\phi(g_1)\phi(g_2)$ for all $g_1,g_2\in G$ (Note that $\phi$ may not be bijective).

This is a weaker condition than isomorphism.

Example

$GL(2,\mathbb{R})=\{A\in M_{2\times 2}(\mathbb{R})|det(A)\neq 0\}$

Then $\phi:GL(2,\mathbb{R})\to (\mathbb{R}-\{0\},\cdot)$ where $\phi(A)=\det(A)$ is a group homomorphism, since $\det(AB)=\det(A)\det(B)$ .

This is not one-to-one but onto, therefore not an isomorphism.

$(\mathbb{Z}_n,+)$ and $D_n$ has homomorphism $(\mathbb{Z}_n,+)\to D_n$ where $\phi(k)=\rho^k$

$\phi(i+j)=\rho^{i+j\mod n}=\rho^i\rho^j=\phi(i)+\phi(j)$ .

This is not onto but one-to-one, therefore not an isomorphism.

Let $G,G'$ be two groups, let $e$ be the identity of $G$ and let $e'$ be the identity of $G'$ .

Let $\phi:G\to G'$ , $\phi(a)=e'$ for all $a\in G$ .

This is a group homomorphism,

\phi(ab)=\phi(a)\phi(b)=e'e'=e'

This is generally not onto and not one-to-one, therefore not an isomorphism.

Corollary for group homomorphism

Let $G,G'$ be groups and $\phi:G\to G'$ be a group homomorphism. $e$ is the identity of $G$ and $e'$ is the identity of $G'$ .

$\phi(e)=e'$
$\phi(a^{-1})=(\phi(a))^{-1}$ for all $a\in G$
If $H\leq G$ , then $\phi(H)\leq G'$ , where $\phi(H)=\{\phi(a)|a\in H\}$ .
If $K\leq G'$ then $\phi^{-1}(K)\leq G$ , where $\phi^{-1}(K)=\{a\in G|\phi(a)\in K\}$ .

Proof

(1) $\phi(e)=e'$

Consider $\phi(ee)=\phi(e)\phi(e)$ , therefore $\phi(e)=e'$ by cancellation on the left.

(2) $\phi(a^{-1})=(\phi(a))^{-1}$

Consider $\phi(a^{-1}a)=\phi(a^{-1})\phi(a)=\phi(e)$ , therefore $\phi(a^{-1})$ is the inverse of $\phi(a)$ in $G'$ .

(3) If $H\leq G$ , then $\phi(H)\leq G'$ , where $\phi(H)=\{\phi(a)|a\in H\}$ .

$e\in H$ implies that $e'=\phi(e)\in\phi(H)$ .
If $x\in \phi(H)$ , then $x=\phi(a)$ for some $a\in H$ . So $x^{-1}=(\phi(x))^{-1}=\phi(x^{-1})\in\phi(H)$ . But $x\in H$ , so $x^{-1}\in H$ , therefore $x^{-1}\in\phi(H)$ .
If $x,y\in \phi(H)$ , then $x,y=\phi(a),\phi(b)$ for some $a,b\in H$ . So $xy=\phi(a)\phi(b)=\phi(ab)\in\phi(H)$ (by homomorphism). Since $ab\in H$ , $xy\in\phi(H)$ .

(4) If $K\leq G'$ then $\phi^{-1}(K)\leq G$ , where $\phi^{-1}(K)=\{a\in G|\phi(a)\in K\}$ .

$e'\in K$ implies that $e=\phi^{-1}(e')\in\phi^{-1}(K)$ .
If $x\in \phi^{-1}(K)$ , then $x=\phi(a)$ for some $a\in G$ . So $x^{-1}=(\phi(x))^{-1}=\phi(x^{-1})\in\phi^{-1}(K)$ . But $x\in G$ , so $x^{-1}\in G$ , therefore $x^{-1}\in\phi^{-1}(K)$ .
If $x,y\in \phi^{-1}(K)$ , then $x,y=\phi(a),\phi(b)$ for some $a,b\in G$ . So $xy=\phi(a)\phi(b)=\phi(ab)\in\phi^{-1}(K)$ (by homomorphism). Since $ab\in G$ , $xy\in\phi^{-1}(K)$ .

Definition for kernel and image of a group homomorphism

Let $G,G'$ be groups and $\phi:G\to G'$ be a group homomorphism.

$\operatorname{ker}(\phi)=\{a\in G|\phi(a)=e'\}=\phi^{-1}(\{e'\})$ is called the kernel of $\phi$ .

Facts:

$\operatorname{ker}(\phi)$ is a subgroup of $G$ . (proof by previous corollary (4))
$\phi$ is onto if and only if $\operatorname{ker}(\phi)=\{e\}$ (the trivial subgroup of $G$ ). (proof forward, by definition of one-to-one; backward, if $\phi(a)=\phi(b)$ , then $\phi(a)\phi(b)^{-1}=e'$ , so $\phi(a)\phi(b^{-1})=e'$ , so $ab^{-1}=e$ , so $a,b=e$ , so $a=b$ )