# Math4302 Modern Algebra (Lecture 19)

## Solution for Midterm I

Not applicable

## Group

Side notes: From previous lecture

What is the group $\mathbb{Z}\times \mathbb{Z}/\langle (a,b)\rangle$, where $a,b\in \mathbb{Z}^+$?

This should be isomorphic to $\mathbb{Z}\times \mathbb{Z}_{\operatorname{gcd}(a,b)}$

<details>
<summary>Proof</summary>

If $\operatorname{gcd}(a,b)=1$, then $\mathbb{Z}\times \mathbb{Z}/\langle (a,b)\rangle\simeq \mathbb{Z}$.

The general isomorphism $\phi:\mathbb{Z}\times \mathbb{Z}/\langle (a,b)\rangle\to \mathbb{Z}\times \mathbb{Z}_{\operatorname{gcd}(a,b)}$ is given by

$$
(a,b)\mapsto (\frac{ma-nb}{d},au+bv\mod d)
$$

where $u,v$ are two integers such that $mu+nv=d$.

Geometrically, partition the lattice to grids then map each element in grid to continuous integer in $ab/\operatorname{gcd}(a,b)=\operatorname{lcm}(a,b)$.

$\phi$ is surjective homomorphism, and $\ker(\phi)=\langle (a,b)\rangle$

</details>

### Center of a group

Recall from previous lecture, the center of a group $G$ is the subgroup of $G$ that contains all elements that commute with all elements in $G$.

$$
Z(G)=\{a\in G\mid \forall g\in G, ag=ga\}
$$

this subgroup is normal and measure the "abelian" for a group.

<details>
<summary>Example</summary>

$Z(S_3)=\{e\}$, all the transpositions are not commutative, so $Z(S_3)=\{e\}$.

---

$Z(GL_n(\mathbb{R}))=\{A\in GL_n(\mathbb{R})\mid AB=BA\text{ for all }B\in GL_n(\mathbb{R})\}$

This is all the multiples of the identity matrix.

Ideas of proof:

1. Consider all matrix that is have non-zero element other than diagonal, we can always find a matrix that don't commute with it.
2. Consider if the diagonal entry are different, the row flip matrix will make your life worse.

Therefore the only choice left the multiple of the identity matrix.

</details>

#### Definition of the commutator of a group

Let $G$ be a group and $a,b\in G$, the commutator $[a,b]$ is defined as $aba^{-1}b^{-1}$.

$[a,b]=e$ if and only if $a$ and $b$ commute.

Some additional properties:

- $[a,b]^{-1}=[b,a]$

#### Definition of commutator subgroup

Let $G'$ be the subgroup of $G$ generated by all commutators of $G$.

$$
G'=\{[a_1,b_1][a_2,b_2]\ldots[a_n,b_n]\mid a_1,a_2,\ldots,a_n,b_1,b_2,\ldots,b_n\in G\}
$$

Then $G'$ is the subgroup of $G$.

- Identity: $[e,e]=e$
- Inverse: $([a_1,b_1],\ldots,[a_n,b_n])^{-1}=[b_n,a_n],\ldots,[b_1,a_1]$

Some additional properties:

- $G$ is abelian if and only if $G'=\{e\}$
- $G'\trianglelefteq G$
- $G/G'$ is abelian
- If $N$ is a normal subgroup of $G$, and $G/N$ is abelian, then $G'\leq N$.

> These implies that $G'$ is the smallest abelian normal subgroup of $G$.

<details>
<summary>Proof</summary>

First we will show that $G'$ is normal

Let $g\in G$ and $h=[a_1,b_1],\ldots,[a_n,b_n]\in G'$, then it is sufficient to show that $ghg^{-1}\in G'$.

Consider arbitrary $[a,b]$

$$
\begin{aligned}
g[a,b]g^{-1}&=ga_1 b_1 a_1^{-1}b_1^{-1}g^{-1}\\
&=ga(g^{-1}g)b(g^{-1} g)a^{-1}(g^{-1}g)b^{-1}g^{-1}\\
&=(gag^{-1})(gbg^{-1})(ga^{-1}g^{-1})(gb^{-1}g^{-1})\\
&=(gag^{-1})(gbg^{-1})(gag^{-1})^{-1}(gbg^{-1})^{-1}\\
&=[gag^{-1},gbg^{-1}]
\end{aligned}
$$

---

Them we will show that $G/G'$ is abelian.

For all $a,b\in G$, we have $[a,b]\in G'$, so $[a,b]^{-1}\in G'$. Therefore $abG'=baG'\iff [b^{-1},a^{-1}]\in G'$. Therefore $G/G'$ is abelian.

</details>
