NoteNextra-origin/content/Math4302/Math4302_L9.md

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.

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(\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.

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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'.

1. \phi(e)=e'
2. \phi(a^{-1})=(\phi(a))^{-1} for all a\in G
3. If H\leq G, then \phi(H)\leq G', where \phi(H)=\{\phi(a)|a\in H\}.
4. 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.

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(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'.

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(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).

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(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)