NoteNextra-origin/content/Math4302/Math4302_L10.md

Math4302 Modern Algebra (Lecture 10)

Groups

Group homomorphism

Recall the kernel of a group homomorphism is the set

\operatorname{ker}(\phi)=\{a\in G|\phi(a)=e'\}

Example

Let \phi:(\mathbb{Z},+)\to (\mathbb{Z}_n,+) where \phi(k)=k\mod n.

The kernel of \phi is the set of all multiples of n.

Theorem for one-to-one group homomorphism

\phi:G\to G' is one-to-one if and only if \operatorname{ker}(\phi)=\{e\}

If \phi is one-to-one, then \phi(G)\leq G', G is isomorphic ot \phi(G) (onto automatically).

If A is a set, then a permutation of A is a bijection f:A\to A.

Cayley's Theorem

Every group G is isomorphic to a subgroup of S_A for some A (and if G is finite then A can be taken to be finite.)

Example

D_n\leq S_n, so A=\{1,2,\cdots,n\}

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\mathbb{Z}_n\leq S_n, (use the set of rotations) so A=\{1,2,\cdots,n\} \phi(i)=\rho^i where i\in \mathbb{Z}_n and \rho\in D_n

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GL(2,\mathbb{R}). Set A=\mathbb{R}^2, for every A\in GL(2,\mathbb{R}), let \phi(A) be the permutation of \mathbb{R}^2 induced by A, so \phi(A)=f_A:\mathbb{R}^2\to \mathbb{R}^2, f_A(\begin{pmatrix}x\\y\end{pmatrix})=A\begin{pmatrix}x\\y\end{pmatrix}

We want to show that this is a group homomorphism.

• \phi(AB)=\phi(A)\phi(B) (it is a homomorphism)

\begin{aligned}
    f_{AB}(\begin{pmatrix}x\\y\end{pmatrix})&=AB\begin{pmatrix}x\\y\end{pmatrix}\\
    &=f_A(B\begin{pmatrix}x\\y\end{pmatrix})\\
    &=f_A(f_B(\begin{pmatrix}x\\y\end{pmatrix}))\\
    &=(f_A\circ f_B)(\begin{pmatrix}x\\y\end{pmatrix})\\
\end{aligned}

• Then we need to show that \phi is one-to-one.

It is sufficient to show that \operatorname{ker}(\phi)=\{e\}.

Solve f_A(\begin{pmatrix}x\\y\end{pmatrix})=\begin{pmatrix}x\\y\end{pmatrix}, the only choice for A is the identity matrix.

Therefore \operatorname{ker}(\phi)=\{e\}.

Proof for Cayley's Theorem

Let A=G, for every g\in G, define \lambda_g:G\to G by \lambda_g(x)=gx.

Then \lambda_g is a permutation of G. (not homomorphism)

• \lambda_g is one-to-one by cancellation on the left.
• \lambda_g is onto since \lambda_g(g^{-1}y)=y for every y\in G.

We claim \phi: G\to S_G define by \phi(g)=\lambda_g is a group homomorphism that is one-to-one.

First we show that \phi is homomorphism.

\forall x\in G

\begin{aligned}
    \phi(g_1)\phi(g_2)&=\lambda_{g_1}(\lambda_{g_2}(x))\\
    &=\lambda_{g_1g_2}(x)\\
    &=\phi(g_1g_2)x\\
\end{aligned}

This is one to one since if \phi(g_1)=\phi(g_2), then \lambda_{g_1}=\lambda_{g_2}\forall x, therefore g_1=g_2.

Odd and even permutations

Definition of transposition

A \sigma\in S_n is a transposition is a two cycle \sigma=(i j)

Fact: Every permutation in S_n can be written as a product of transpositions. (may not be disjoint transpositions)

Example of a product of transpositions

Consider (1234)=(14)(13)(12).

In general, (i_1,i_2,\cdots,i_m)=(i_1i_m)(i_2i_{m-1})(i_3i_{m-2})\cdots(i_1i_2)

This is not the unique way.

(12)(34)=(42)(34)(23)(12)

But the parity of the number of transpositions is unique.

Theorem for parity of transpositions

If \sigma\in S_n is written as a product of transposition, then the number of transpositions is either always odd or even.

Definition of odd and even permutations

\sigma is an even permutation if the number of transpositions is even.

\sigma is an odd permutation if the number of transpositions is odd.