NoteNextra-origin/content/Math4302/Math4302_L4.md

Math4302 Modern Algebra (Lecture 4)

Groups

Group Isomorphism

Definition of isomorphism

Let (G_1,*_1) and (G_2,*_2) be two groups. Then (G_1,*_1) and (G_2,*_2) are isomorphic if there exists a bijection f:G_1\to G_2 such that for all x,y\in G_1, f(x*y)=f(x)*f(y). We say that (G_1,*_1) is isomorphic to (G_2,*_2).

(G_1,*_1)\simeq (G_2,*_2)

Example and non-example for isomorphism

As we have seen in class, (\mathbb{Z}_4,+) and (\{1,-1,i,-i\},*) are isomorphic.

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(\mathbb{Z},+) and (\mathbb{R},+) are not isomorphic. There is no bijection from (\mathbb{Z},+) to (\mathbb{R},+).

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Let M_2(\mathbb{R}) denotes the set of 2\times 2 matrices with addition. Then (\mathbb{R}^4,+) and (M_2(\mathbb{R}),+) are isomorphic.

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(\mathbb{Z},+) and (\mathbb{Q},+) are not isomorphic.

• There exists bijection mapping \mathbb{Z}\to \mathbb{Q}, but

Suppose we have f(1)=a\in \mathbb{Q}, so there exists unique element f(x), x\in \mathbb{Z} such that f(x)=\frac{a}{2}, if such function f is isomorphic (preserves addition), then f(2x)=f(x)+f(x)=a. So 2x=1, such x does not exist in \mathbb{Z}.

Isomorphism of Groups defines an equivalence relation

Isomorphism of groups is an equivalence relation.

1. Reflexive: (G_1,*_1)\simeq (G_1,*_1)
2. Symmetric: (G_1,*_1)\simeq (G_2,*_2)\implies (G_2,*_2)\simeq (G_1,*_1)
3. Transitive: (G_1,*_1)\simeq (G_2,*_2)\land (G_2,*_2)\simeq (G_3,*_3)\implies (G_1,*_1)\simeq (G_3,*_3)

Easy to prove using bijective maps and definition of isomorphism.

Some fun facts

For any prime number, there is only one group of order p for any p\in\mathbb{N}.

OEIS A000001

Example of non-abelian finite groups

Permutations (Symmetric groups) S_n.

Let A be a set of n elements, a permutation of A is a bijection from A to A.

\sigma: A\to A

Let A be a finite set, A=\{1,2,...,n\}. Then there are n! permutations of A.

We can denote each permutation on A=\{1,2,...,n\} by

\sigma=\begin{pmatrix}
1&2&...&n\\
\sigma(1)&\sigma(2)&...&\sigma(n)
\end{pmatrix}

Symmetric Groups

The set of permutation on a set A form a group under function composition.

• Identity: $\sigma_{id}=\begin{pmatrix} 1&2&...&n\ 1&2&...&n \end{pmatrix}$
• Inversion: If f: A\to A is a bijection, then f^{-1}: A\to A is a bijection and is the inverse of f.
• Associativity: (\sigma_1*\sigma_2)*\sigma_3=\sigma_1*(\sigma_2*\sigma_3)

|S_n|=n!

When n=1,2, the group is abelian.

but when n=3, we have some \sigma,\tau\in S_3 such that \sigma*\tau\neq \tau*\sigma.

Let $\sigma=\begin{pmatrix} 1&2&3\ 2&3&1 \end{pmatrix}$ and $\tau=\begin{pmatrix} 1&2&3\ 3&2&1\ \end{pmatrix}$, then $\sigma*\tau=\begin{pmatrix} 1&2&3\ 1&3&2 \end{pmatrix}$ and $\tau*\sigma=\begin{pmatrix} 1&2&3\ 2&1&3 \end{pmatrix}$.

Therefore \tau*\sigma\neq \sigma*\tau.

Then we have a group of order 3!=6 that is not abelian.

For any n\geq 3, S_n is not abelian. (Proof by induction, keep \sigma,\tau extra entries being the same$).

Another notation for permutations is using the cycle.

Suppose we have $\sigma=\begin{pmatrix} 1&2&3&4\ 2&3&1&4\ \end{pmatrix}$, then we have the cycle (1,2,3)(4).

this means we send 1\to 2\to 3\to 1 and 4\to 4.

Some case we ignore (4) and just write (1,2,3).

Tip

From now on, we use G to denote (G,*) and ab to denote a*b to save chalks.

If G is abelian, we use + to denote the group operations

• Instead of a*b or ab, we write a+b.
• Instead of a^{-1}, we write -a.
• Instead of e, we write 0.
• Instead of a^{n}, we write na.