Algebraic Structures Coursework

Question C1.1 (55 total marks)

Considering the multiplicative group $GL(2,\mathbb{R})$ and its subset $H$, such that…

$$GL(2,\mathbb{R})=\left\{\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]:a,b,c,d\in \mathbb{R},ad-bc\ne 0\right\}$$ $$\text{and }H=\left\{\left[\begin{array}{cc}a& b\\ 0& d\end{array}\right]:a,b,d\in \mathbb{R},ad=1\right\}$$

Part A (15 marks)

Let $A=\left[\begin{array}{cc}{a}_1& b_1\\ 0& d_1\end{array}\right]$ and $B=\left[\begin{array}{cc}{a}_2& b_2\\ 0& d_2\end{array}\right]$ be in $H$. Then, from the One-Step Subgroup Test, we can show that the non-empty subset $H$ is a subgroup of $GL(2,\mathbb{R})$ over matrix multiplication if…

$$\begin{array}{rl}A{B}^{-1}& =\left[\begin{array}{cc}{a}_1& b_1\\ 0& d_1\end{array}\right]{\left[\begin{array}{cc}{a}_2& b_2\\ 0& d_2\end{array}\right]}^{-1}\\ & =\left[\begin{array}{cc}{a}_1& b_1\\ 0& d_1\end{array}\right]\left[\begin{array}{cc}\frac{d_2}{a_2{d}_2}& \frac{-b_2}{a_2{d}_2}\\ 0& \frac{a_2}{a_2{d}_2}\end{array}\right]\\ & =\left[\begin{array}{cc}{a}_1& b_1\\ 0& d_1\end{array}\right]\left[\begin{array}{cc}\frac{1}{a_2}& -\frac{b_2}{a_2{d}_2}\\ 0& \frac{1}{d_2}\end{array}\right]\\ & =\left[\begin{array}{cc}\frac{a_1}{a_2}& -\frac{a_1{b}_2}{a_2{d}_2}+\frac{b_1}{d_2}\\ 0& \frac{d_1}{d_2}\end{array}\right]\\ & =\left[\begin{array}{cc}\frac{a_1}{a_2}& \frac{a_2{b}_1-a_1{b}_2}{a_2{d}_2}\\ 0& \frac{d_1}{d_2}\end{array}\right]:\frac{a_1}{a_2}\cdot \frac{d_1}{d_2}=\frac{a_1{d}_1}{a_2{d}_2}=\frac{1}{1}\text{ by defintion}\\ & \in H\end{array}$$

Therefore, because $H$ is non-empty and for all $A,B$ then $A{B}^{-1}\in H$, we conclude that $H$ is a subgroup of $GL(2,\mathbb{R})$. $\square$

Part B (10 marks)

Let $P=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\in H$ such that the mapping $f:H\to H$ is given by $f(A)=P^{-1}AP$ for each $A\in H$.

The mapping $f$ is a group homomorphism, since it preserves the group operation:

$$\boxed{\begin{array}{rl}\forall A,B\in H,f(AB)& =P^{-1}(AB)P\\ & =(P^{-1}AP)(P^{-1}BP)\text{ by associativity}\\ & =f(A)f(B)\end{array}}$$

Part C (10 marks)

The aforementioned mapping is also a group isomorphism, as it’s both injective and surjective:

First, we can check injectivity by assuming $f(A)=f(B)$ for $A,B\in H$, such that…

$$P^{-1}AP=P^{-1}BP⟺A=B\therefore f\text{ is injective}$$

Second, we can check surjectivity by finding $A\in H:f(A)=C\in H$…

$$\begin{array}{rl}f(A)& =P^{-1}AP\\ & =\left[\begin{array}{cc}1& -2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}{a}_1& b_1\\ 0& d_1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\\ & =\left[\begin{array}{cc}1& -2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}{a}_1& 2a_1+b_1\\ 0& d_1\end{array}\right]\\ & =\left[\begin{array}{cc}{a}_1& 2a_1+b_1-2d_1\\ 0& d_1\end{array}\right]:a_1{d}_1=1\text{ by definition}\\ & \in H\therefore f\text{ is surjective}\end{array}$$ $$\boxed{\therefore \text{the mapping is a group isomorphism}}$$

Part D (10 marks)

The subgroup $H$ isn’t an abelian subgroup because matrix multiplication is not commutative:

$$\begin{array}{rl}AB=BA& ⟹\left[\begin{array}{cc}{a}_1{a}_2& a_1{b}_2+b_1{d}_1\\ 0& d_1{d}_2\end{array}\right]=\left[\begin{array}{cc}{a}_2{a}_1& a_2{b}_1+b_2{d}_2\\ 0& d_2{d}_1\end{array}\right]\\ & ⟹a_1{b}_2+b_1{d}_1=a_2{b}_1+b_2{d}_2\\ & ⟺a_1=a_2,b_1=b_2,d_1=d_2\\ & \therefore \text{only commutative if }A=B\text{, hence not generally abelian.}\end{array}$$

Part E (10 marks)

Let the subgroup $K=\left\{\left[\begin{array}{cc}1& b\\ 0& 1\end{array}\right]:b\in \mathbb{R}\right\}$. Given that $K$ is a subgroup of $H$, $K$ is a normal subgroup if $\forall k\in K,h\in H$ the conjugate $hk{h}^{-1}$ is also in $K$.

Let $h=\left[\begin{array}{cc}a& b\\ 0& d\end{array}\right]\in H$ (where $ad=1$) and $k=\left[\begin{array}{cc}1& b^{\prime }\\ 0& 1\end{array}\right]\in K$…

$$\begin{array}{rl}hk{h}^{-1}& =\left[\begin{array}{cc}a& b\\ 0& d\end{array}\right]\left[\begin{array}{cc}1& b^{\prime }\\ 0& 1\end{array}\right]\left[\begin{array}{cc}\frac{1}{a}& \frac{-b}{ad}\\ 0& \frac{1}{d}\end{array}\right]\\ & =\left[\begin{array}{cc}a& b\\ 0& d\end{array}\right]\left[\begin{array}{cc}\frac{1}{a}& \frac{-b+a{b}^{\prime }}{ad}\\ 0& \frac{1}{d}\end{array}\right]\\ & =\left[\begin{array}{cc}1& \frac{a{b}^{\prime }}{d}\\ 0& 1\end{array}\right]:\frac{a{b}^{\prime }}{d}=\frac{a}{d}{b}^{\prime }\end{array}$$

Whilst $b^{\prime }\in \mathbb{R}$, $\frac{a}{d}$ is not always in $\mathbb{R}$, therefore the conjugate $hk{h}^{-1}$ is not always in $K$:

$$\boxed{\therefore K\text{ is not a normal subgroup of }H}$$

Question C1.2 (45 total marks)

Considering the group ${\mathbb{Z}}_{12}$ of integers $\text{mod }12$…

Part A (10 marks)

To find the order of the element $\bar{4}$ in the group ${\mathbb{Z}}_{12}$, we need to solve…

$$n\cdot \bar{4}=\bar{0}\text{ in }{\mathbb{Z}}_{12}⟺4n\equiv 0\text{ mod }12⟺n=3\therefore \boxed{\text{the order is 3}}$$

Part B (20 total marks)

Part I (10 marks)

To find the subgroup $⟨\bar{2}⟩$ of ${\mathbb{Z}}_{12}$, we compute the multiples of $\bar{2}$:

$$\boxed{\begin{array}{rl}⟨\bar{2}⟩& =\{n\cdot \bar{2}:n\in \mathbb{Z}\}\\ & =\{\bar{0},\bar{2},\bar{4},\bar{6},\bar{8},\overline{10}\}\end{array}}$$

Part II (10 marks)

To find the subgroup $⟨\bar{6}⟩$ of ${\mathbb{Z}}_{12}$, we compute the multiples of $\bar{6}$:

$$\boxed{\begin{array}{rl}⟨\bar{6}⟩& =\{n\cdot \bar{6}:n\in \mathbb{Z}\}\\ & =\{\bar{0},\bar{6}\}\end{array}}$$

Part C (15 marks)

We can show that the mapping $h:{\mathbb{Z}}_{12}\to {\mathbb{Z}}_{12}$, given by $h(x)=\bar{2}x$, is a group homomorphism by showing that it preserves the group operation:

$$\begin{array}{rl}h(a+b)& =\bar{2}(a+b):a,b\in {\mathbb{Z}}_{12}\\ & =\bar{2}a+\bar{2}b\\ & =h(a)+h(b)\therefore \boxed{h\text{ is a group homomorphism}}\end{array}$$

We can then find its kernel by definition:

$$\boxed{\begin{array}{rl}\ker (h)& =\{x\in {\mathbb{Z}}_{12}:h(x)=\bar{0}\}\\ & =\{\bar{2}x=\bar{0}\}\\ & =\{2x\equiv 0\text{ mod }12\}\\ & =\{x\equiv 0\text{ mod }6\}\\ & =\{\bar{0},\bar{6}\}\end{array}}$$