🏠

Groups and group actions problem sheet 1

 
  • Feb 01, 2022
  1. For each of the following sets $S$ and binary operations ∗ on $S$, state whether (a) ∗ is associative, (b) ∗ is commutative, (c) an identity exists, (d) inverses exist.
    (i) $S=\mathbb N, m ∗ n =\max\{m,n\}$.
    (ii) $S=\mathbb Z, m ∗ n = m + n + 1$.
    (iii) $S=M_n(\mathbb R)$ where $n > 2$ and $A ∗ B =\frac12(AB + BA)$.
    (iv) $S=\{f :\mathbb R →\mathbb R\}, f ∗ g = f ◦ g$.
    Solution.
    (i)associative, commutative, 0 or 1, no inverse($\max\{0,2\}=\max\{1,2\}$, so 2 doesn't have inverse).
    (ii)associative, commutative, identity is -1, inverse of $n$ is $-n-2$.
    (iii)not associative$\Big[A=\begin{pmatrix}0&1\\-1&0\end{pmatrix},B=\begin{pmatrix}-1&0\\0&1\end{pmatrix},A(AB)=O_2\ne-B=(AA)B\Big]$, commutative, identity=$I_n$, no inverse if $A$ is singular.
    (iv)associative, not commutative, identity is $f(x)=x$, non-injective functions don't have inverse.
  2. There are 56 different Latin squares of size 5×5 whose first row and first column are both $(e, a, b, c, d)$. Construct one such square which is a group table, justifying your reasons for it representing a group, and one such square that does not represent a group, again saying why.
    Solution.
    $\begin{array}{c|ccccc}
    \mathbb Z_5^+&a&b&c&d&e\\\hline
    a&a&b&c&d&e\\
    b&b&c&d&e&a\\
    c&c&d&e&a&b\\
    d&d&e&a&b&c\\
    e&e&a&b&c&d\end{array}$ is a group table by Proposition 130 (a).
    $\begin{array}{c|ccccc}&a&b&c&d&e\\\hline a&a&b&c&d&e\\b&b&a&e&c&d\\c&c&d&b&e&a\\d&d&e&a&b&c\\e&e&c&d&a&b\end{array}$ is not a group table, since $(bc)d=a≠d=b(cd)$.
  3. Let $A$ and $B$ be complex $n×n$ matrices. If $A=(a_{ij})$ then we define its complex conjugate as $\overline A = (\overline{a_{ij}})$. Show that $\overline{A+B}=\overline{A}+\overline{B}, \overline{AB}=\overline{A}\ \overline{B}$. Show that $U(n)$ is a group. Show further that $U(1)$ is Abelian and that $U(n)$ is non-Abelian for $n≥2$.
    Solution.
    $\overline{A+B}=((A+B)^*)^{\sf T}=(A^*+B^*)^{\sf T}=(A^*)^{\sf T}+(B^*)^{\sf T}=\overline{A}+\overline{B}$,$\overline{AB}=((AB)^*)^{\sf T}=(B^*A^*)^{\sf T}=(A^*)^{\sf T}(B^*)^{\sf T}=\overline{A}\ \overline{B}$.
    $I∈U(n)$. $∀A∈U(n),A^{-1}=\overline{A^T}=(\overline{(A^{-1})^T})^{-1}$, so $A^{-1}∈U(n)$.
    ∀$A,B∈U(n)$,$AB=(\overline{A^{\sf T}})^{-1}(\overline{B^{\sf T}})^{-1}=(\overline{B^{\sf T}}\overline{A^{\sf T}})^{-1}=(\overline{B^{\sf T}A^{\sf T}})^{-1}=(\overline{(AB)^{\sf T}})^{-1}⇒AB∈U(n)$.
    Therefore $U(n)$ is a subgroup of $GL(n)$.
    An element of $U(1)$ is a complex 1×1 matrix, so $U(1)$ is Abelian.
    For $n≥2$, take $U=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ and $V=\begin{pmatrix}0&1\\1&0\end{pmatrix}$. We have $\begin{pmatrix}U&\\&I_{n-2}\end{pmatrix}\begin{pmatrix}V&\\&I_{n-2}\end{pmatrix}=\begin{pmatrix}0&1\\-1&0\\&&1_{n-2}\end{pmatrix}≠\begin{pmatrix}0&-1\\1&0\\&&I_{n-2}\end{pmatrix}=\begin{pmatrix}V&\\&I_{n-2}\end{pmatrix}\begin{pmatrix}U&\\&I_{n-2}\end{pmatrix}$.
  4. An affine transformation of $\mathbb R^2$ is one of the form $\left(\begin{array}{l}x \\ y\end{array}\right) \mapsto A\left(\begin{array}{l}x \\ y\end{array}\right)+\mathbf{b}$ where $A$ is an invertible 2×2 matrix and $\bf b$ is a 2×1 column vector. Let $g_1$ and $g_2$ be affine transformations of $\mathbb R^2$. Show that their composition $g_2◦g_1$ is an affine transformation. Show further that the affine transformations of $\mathbb R^2$ form a group $AGL(2,\mathbb R)$ under composition.
    Proof.
    Let $g_1$ and $g_2$ be affine transformations on $\mathbb R^2$, their composition $g_2∘g_1$ has form $\left(\begin{array}{l}x \\ y\end{array}\right) \mapsto A_2\left(A_1\left(\begin{array}{l}x \\ y\end{array}\right)+\mathbf{b}_1\right)+\mathbf b_2=A_2A_1\left(\begin{array}{l}x \\ y\end{array}\right)+A_2\mathbf{b}_1+\mathbf b_2$, so $g_2∘g_1$ is an affine transformation.
    Composition of transformation is associative. The identity is $A=I,\mathbf b=0$.
    The inverse of $\left(\begin{array}{l}x \\ y\end{array}\right) \mapsto A\left(\begin{array}{l}x \\ y\end{array}\right)+\mathbf{b}$ is $\left(\begin{array}{l}x \\ y\end{array}\right) \mapsto A^{-1}\left(\begin{array}{l}x \\ y\end{array}\right)-A^{-1}\mathbf{b}$.
    Remark. You can show geometrically that every affine transformation can be decomposed (uniquely) as a composite of a linear map fixing the origin followed by a translation. $AGL(n,\mathbb R)$ is the semi-direct product of $\mathbb R^n$ and $GL(n,\mathbb R)$. In fact, the subgroup of translations(isomorphic to $\mathbb R^n$) is a normal subgroup of $AGL(n,\mathbb R)$(see Example 225 in the lecture notes) but $GL(n,\mathbb R)$ is not normal(Let $f(x) = Ax +\mathbf b, A ∈ GL(n,\mathbb R)$ and $g(x)=Bx,B∈GL(n,\mathbb R)$, $f^{-1}gf(x)=A^{-1}(B(Ax+\mathbf b)-\mathbf b)$ is not necessarily equal to $g(x)=Bx$, so $GL(n,\mathbb R)$ is not normal subgroup of $AGL(n,\mathbb R)$ by definition)(see this post)
  5. The following Cayley table describes a group $G$. (Dihedral group $D_8$)\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline ∗&e&a&b&c&d&f&g&h\\\hline e&e&a&b&c&d&f&g&h\\\hline a&a&e&h&g&f&d&c&b\\\hline b&b&c&d&f&g&h&e&a\\\hline c&c&b&a&e&h&g&f&d\\\hline d&d&f&g&h&e&a&b&c\\\hline f&f&d&c&b&a&e&h&g\\\hline g&g&h&e&a&b&c&d&f\\\hline h&h&g&f&d&c&b&a&e\\\hline \end{array}(i) Find the inverse of each element of $G$.
    (ii) Are there any elements of $G$, other than $e$, which commute with every element of $G$?
    (iii) Determine the order of each element of $G$.
    (iv) Show that $G=\left\{e,b,b^2,b^3,a,ba,b^2a,b^3a\right\}$ and that $ab = b^3a$.
    (v) In all, $G$ has ten subgroups, of orders 1, 2, 2, 2, 2, 2, 4, 4, 4, 8. List the subgroups of $G$.
    Solution.
    (i) $a^{-1}=a,b^{-1}=g,c^{-1}=c,d^{-1}=d,f^{-1}=f,g^{-1}=b,h^{-1}=h$
    (ii) $d$ commutes with every element of $G$.
    (iii) $\DeclareMathOperator{\ord}{ord}\ord e=\ord a=\ord c=\ord d=\ord f=\ord h=2,$$\ord b=\ord g=4.$
    (iv)$e,b,b^2=d,b^3=g,a,ba=c,b^2a=f,b^3a=h$ are distinct, so $G=\left\{e,b,b^2,b^3,a,ba,b^2a,b^3a\right\}$. $ab=h=b^3a$.
    (v)$\{e\},\{e,a\},\{e,c\},\{e,d\},\{e,f\},\{e,h\},$$\{e,a,d,f\},\{e,b,d,g\},\{e,c,d,h\},\{e,a,b,c,d,f,g,h\}$
  6. (i) Let $G, H$ be groups. Show that $G × H$ is Abelian if and only if $G, H$ are both Abelian.
    (ii) Show that the map $\phi: \mathbb{C}^{*} \rightarrow(0, \infty) \times S^{1}$ given by $\phi(z)=(|z|, z /|z|)$ is an isomorphism.
    (iii) Show that $S^1$ is isomorphic to $SO(2)$ but that $S^1×\{±1\}$ is not isomorphic to $O(2)$.
    Proof.
    (i) ⇒:If $G×H$ is Abelian, $(g_1,e)(g_2,e)=(g_2,e)(g_1,e)$⇒$(g_1g_2,e)=(g_2g_1,e)$⇒$G$ is Abelian. Likewise, $H$ is Abelian.
    ⇐:If $G,H$ are both Abelian, $(g_1,h_1)(g_2,h_2)=(g_1g_2,h_1h_2)$$=(g_2g_1,h_2h_1)=(g_2,h_2)(g_1,h_1)$ for any $g_1,g_2∈G,h_1,h_2∈H$.
    (ii) $∀z_1,z_2∈\mathbb C^*$, $\phi(z_1)\phi(z_2)=(|z_1|,z_1/|z_1|)(|z_2|,z_2/|z_2|)=(|z_1|·|z_2|,z_1z_2/(|z_1|·|z_2|))=(|z_1z_2|,z_1z_2/|z_1z_2|)=\phi(z_1z_2)$, so $\phi$ is a homomorphism.
    $\phi^{-1}(r,u)=ru$, so $\phi$ is invertible. Hence $\phi$ is an isomorphism.
    (iii) For $M∈SO(2),$ $M$ has the form $M=\begin{pmatrix}\cosθ&-\sinθ\\\sinθ&\cosθ\end{pmatrix}$, let $ϕ(M)=\cosθ+i\sinθ$, it is a bijection from $SO(2)$ to $S^1$, and $M_1M_2=\begin{pmatrix}\cosθ_1&-\sinθ_1\\\sinθ_1&\cosθ_1\end{pmatrix}\begin{pmatrix}\cosθ_2&-\sinθ_2\\\sinθ_2&\cosθ_2\end{pmatrix}=\begin{pmatrix}\cos(θ_1+θ_2)&-\sin(θ_1+θ_2)\\\sin(θ_1+θ_2)&\cos(θ_1+θ_2)\end{pmatrix}$, $ϕ(M_1)ϕ(M_2)=(\cosθ_1+i\sinθ_1)(\cosθ_2+i\sinθ_2)$$=\cosθ_1\cosθ_2-\sinθ_1\sinθ_2+i(\sinθ_1\cosθ_2+\cosθ_1\sinθ_2)=\cos(θ_1+θ_2)+i\sin(θ_1+θ_2)$, therefore $ϕ(M_1)ϕ(M_2)=ϕ(M_1M_2)$, so ϕ is an isomorphism from $SO(2)$ to $S^1$.
    $S^1×\{±1\}$ is Abelian but $O(2)$ is not, so they aren't isomorphic. [Or use the fact that $O(2)$ contains infinitely many elements of order 2 (all the reflections) but $S^1×\{±1\}$ contains only 3:$((-1,0),1),((-1,0),-1),((1,0),-1)$.]

    Starter
    S1. Show that the general linear group $GL_n(\mathbb R)$ (of invertible $n×n$ real matrices) does indeed form a group under matrix multiplication.
    Solution.
    • closed under matrix multiplication: if $A$ and $B$ are invertible, then we know (from Linear Algebra I) that $A B$ is invertible.
    • identity: $I \in G L_{n}(\mathbb{R})$.
    • associativity: matrix multiplication is associative.
    • inverses: if $A \in G L_{n}(\mathbb{R})$ then $A$ is invertible. Then $A A^{-1}=I=A^{-1} A$, so also $A^{-1}$ is invertible and hence $A^{-1} \in G L_{n}(\mathbb{R})$.
    S2. Using the notation from the lecture notes, show that $e, r, r^{2}, r^{3}, s, r s, r^{2} s, r^{3} s$ are 8 distinct symmetries of the square.
    Solution.
    As with the example for the equilateral triangle in the notes, for each symmetry we see how it permutes the vertices.\begin{align*} e &:\left(\begin{array}{llll}1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4\end{array}\right) \\ r &:\left(\begin{array}{llll}1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1\end{array}\right) \\ r^{2} &:\left(\begin{array}{llll}1 & 2 & 3 & 4 \\ 3 & 4 & 1 & 2\end{array}\right) \\ r^{3} &:\left(\begin{array}{llll}1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3\end{array}\right) \\ s &:\left(\begin{array}{llll}1 & 2 & 3 & 4 \\ 2 & 1 & 4 & 3\end{array}\right) \\ r s &:\left(\begin{array}{llll}1 & 2 & 3 & 4 \\ 3 & 2 & 1 & 4\end{array}\right) \\ r^{2} s &:\left(\begin{array}{llll}1 & 2 & 3 & 4 \\ 4 & 3 & 2 & 1\end{array}\right) \\ r^{3} s &:\left(\begin{array}{llll}1 & 2 & 3 & 4 \\ 1 & 4 & 3 & 2\end{array}\right) \end{align*}These are 8 distinct permutations, so the 8 symmetries are distinct.
    S3. Show that the product operation ∗ on $G × H$ is associative.
    Solution.
    Take $\left(g_{1}, h_{1}\right),\left(g_{2}, h_{2}\right),\left(g_{3}, h_{3}\right) \in G \times H$. Then\begin{align*}
    \left(\left(g_{1}, h_{1}\right) *\left(g_{2}, h_{2}\right)\right) *\left(g_{3}, h_{3}\right) &=\left(g_{1} *_{G} g_{2}, h_{1} *_{H} h_{2}\right) *\left(g_{3}, h_{3}\right) \\
    &\left.=\left(\left(g_{1} *_{G} g_{2}\right) *_{G} g_{3},\left(h_{1} *_{H} h_{2}\right) *_{H} h_{3}\right)\right) \\
    &=\left(g_{1} *_{G}\left(g_{2} *_{G} g_{3}\right), h_{1} *_{H}\left(h_{2} *_{H} h_{3}\right)\right)\quad \text { as } *_{G} \text { and } *_{H} \text { associative } \\
    &=\left(g_{1}, h_{1}\right) *\left(g_{2} *_{G} g_{3}, h_{2} *_{H} h_{3}\right) \\
    &=\left(g_{1}, h_{1}\right) *\left(\left(g_{2}, h_{2}\right) *\left(g_{3}, h_{3}\right)\right)
    \end{align*}so $*$ is associative.

    Pudding
    P1. Let $G$ be a finite group of even order. Must it contain an element of order 2?
    Solution. (This is Corollary 162 in lecture notes.)
    Yes. Suppose that $G$ doesn't contain an element of order 2, for any $a∈G\setminus\{e\}$, we can pair it up with $a^{-1}$, so the order of $G$ must be odd.
    (In general, Cauchy's theorem says that if a prime $p$ divides $|G|$ then $G$ has an element of order $p$. The case $p=2$ has the particularly easy proof given above.)
    P2.
    (i) Is there a group in which no non-identity element is its own inverse?
    (ii) Is there a group in which every non-identity element is its own inverse?
    Solution.
    (i) Yes. For example, consider $C_5$. Here every non-identity element has order 4, and hence is not its own inverse.
    (ii) Yes. For example, consider $C_2$. In fact there are many more examples. We’ll return to groups in which every non-identity element is its own inverse on a future sheet.
    P3. How many ways are there to complete the following grid so that it is the Cayley table of a group?\begin{array}{c|ccc}∗&e&a&b&c\\\hline e&&&&\\a&&&&\\b&&&&\\c&&&&\end{array}Solution. We know what the first row and column must be, so we can fill in some entries immediately.\begin{array}{c|cccc}
    * & e & a & b & c \\
    \hline e & e & a & b & c \\
    a & a & & & \\
    b & b & & & \\
    c & c & & &
    \end{array}There are then three possibilities for $a * a$, namely $b, c$ and $e$ (note that it cannot be $a$, because each element must appear exactly once in each row and column).
    Case 1    : $a * a=b$. Then we find the remaining entries are all determined (using the 'every element in every row and column' property), and we get the table\begin{array}{c|cccc}
    * & e & a & b & c \\
    \hline e & e & a & b & c \\
    a & a & b & c & e \\
    b & b & c & e & a \\
    c & c & e & a & b
    \end{array}This does correspond to a group table. Perhaps the easiest way to see this is to note that it's the group table of $C_{4}$ (with $a^{2}=b$ and $a^{3}=c$ ) - this is probably more convenient than checking associativity by hand.
    Case 2: $a * a=c$. As in Case 1 , it turns out that there are no more decisions to make. We get the table\begin{array}{c|cccc}
    * & e & a & b & c \\
    \hline e & e & a & b & c \\
    a & a & c & e & b \\
    b & b & e & c & a \\
    c & c & b & a & e
    \end{array}which again corresponds to $C_{4}$, this time with $a^{2}=c$ and $a^{3}=b$.
    Case 3: $a * a=e$. Then we can fill in some entries, but when we reach\begin{array}{c|cccc}* & e & a & b & c \\
    \hline e & e & a & b & c \\
    a & a & e & c & b \\
    b & b & c & & \\
    c & c & b & &\end{array}we have to make another decision.
    Case 3a: $a * a=e$ and $b * b=e$. Then everything else is determined (there is no more choice), and we get\begin{array}{c|cccc}* & e & a & b & c \\
    \hline e & e & a & b & c \\
    a & a & e & c & b \\
    b & b & c & e & a \\
    c & c & b & a & e\end{array}which corresponds to $C_{2} \times C_{2}$.
    Case 3b: $a * a=e$ and $b * b=a$. Then again everything else is determined, leading to\begin{array}{c|cccc}
    * & e & a & b & c \\
    \hline e & e & a & b & c \\
    a & a & e & c & b \\
    b & b & c & a & e \\
    c & c & b & e & a\end{array}which is the group table of $C_{4}$, with $b$ as a generator and $b^{2}=a, b^{3}=c$.
    So there are just two possible groups of order 4, up to isomorphism, and they are both Abelian.