Math 120A HW 8 Solutions

  1. Let $\phi : G \to G'$ be a homomorphism of groups.
    1. Denote the identity element of $G'$ by $e'$. If $h \in \ker(\phi)$ and $g \in G$, then $\phi(ghg^{-1}) = \phi(g)\phi(h)\phi(g)^{-1} = \phi(g)e'\phi(g)^{-1} = \phi(g)\phi(g)^{-1} = e'$, so $ghg^{-1} \in \ker(\phi)$.
    2. If $G'$ is abelian and $a,b \in G$ then $\phi(aba^{-1}b^{-1}) = \phi(a)\phi(b)\phi(a)^{-1}\phi(b)^{-1} = \phi(a)\phi(a)^{-1}\phi(b)\phi(b)^{-1} = e'$, so $aba^{-1}b^{-1} \in \ker(\phi)$.
  2. Recall that $SL_2(\mathbb{R})$ is the group of $2 \times 2$ real matrices of determinant $1$, and recall that it is a normal subgroup of $GL_2(\mathbb{R})$.
    1. The determinant of $g$ is $-1$, which is nonzero, so $g \in GL_2(\mathbb{R})$. The determinant of $h$ is $1$, so $h \in SL_2(\mathbb{R})$. We have $$ gh = \begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix} \ne \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} = hg.$$
    2. We have $$ gh = h'g \iff h' = ghg^{-1} = \begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}.$$
  3. Let $H$ be the cyclic subgroup of $\mathbb{Z} \times \mathbb{Z}$ generated by $(1,1)$.
    1. $(i,j) + H = (k,l) + H \iff (i,j) - (k,l) \in H \iff i-k = j-l$.
    2. Consider an arbitrary element $(i,j) + H \in (\mathbb{Z} \times \mathbb{Z}) / H$. We have $(i,j) + H = (0,j-i) + H = (j-i)((0,1)+H) \in \langle (0,1)+H \rangle$.
    3. We have $n((0,1) + H) = (0,n) + H$. By part a, this is equal to the identity element $(0,0) + H$ of $(\mathbb{Z} \times \mathbb{Z}) / H$ if and only if $n=0$.
  4. Consider an arbitrary element of $(ab)H$, say $abh$ where $h \in H$. Denoting the identity element of $G$ by $e$, we have $abh = (ae)(bh) = cd$ where $c = ae \in aH$ and $d = bh \in bH$. Conversely, consider $cd$ where $c \in aH$ and $d \in bH$. Say $c = ah_1$ and $d = bh_2$ where $h_1, h_2 \in H$. Then we have $cd = ah_1bh_2 = abh_1'h_2$ for some $h_1' \in H$, and this is in $(ab)H$. (We used normality to write $h_1b$ as $bh_1'$ with $h_1' \in H$.)
  5. Let $G$ be a group. For $x,y \in G$ we say that $y$ is conjugate to $x$ if $y = i_g(x)$ for some $g \in G$, where the inner automorphism $i_g$ of $G$ is defined by $i_g(x) = gxg^{-1}$.
    1. For reflexivity, let $x \in G$. Note that $i_e(x) = exe^{-1} = exe = x$ where $e$ denotes the identity element of $G$, so $x$ is conjugate to itself. For symmetry, let $x,y \in G$ and suppose that $y = i_g(x) = gxg^{-1}$ where $g \in G$. Multiplying on the right by $g$ and on the left by $g^{-1}$, we have $g^{-1}yg = x$, so $x = i_{g^{-1}}(y)$. For transitivity, let $x,y,z \in G$ and suppose that $y = i_g(x)$ and $z = i_h(y)$ where $g,h \in G$. Then $z = i_h(i_g(x)) = i_h(gxg^{-1}) = hgxg^{-1}h^{-1} = (hg)x(hg)^{-1} = i_{hg}(x)$.
    2. $\{\text{identity}\}, \{(1\;2),(1\;3),(2\;3)\}, \{(1\;2\;3),(1\;3\;2)\}$.
  6. Define $$h = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} \text{ and } g = \begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}.$$ Then $h \in H$ and $g \in GL_2(\mathbb{R})$, but we have $$ghg^{-1} = \begin{pmatrix} 1 & -2\\ 0 & -1 \end{pmatrix} \notin H.$$