Math 120A HW 9 Solutions

1. For this problem it will be helpful to use the fundamental theorem of finite abelian groups.
   1. $\mathbb{Z}_4 \times \mathbb{Z}_9 \times \mathbb{Z}_7$, $\mathbb{Z}_4 \times \mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_7$, $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_9 \times \mathbb{Z}_7$, and $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_7$,
   2. It suffices to show that each of the listed groups has an element of order $42$, because this is a structural property. The order of $(2,3,1)$ in $\mathbb{Z}_4 \times \mathbb{Z}_9 \times \mathbb{Z}_7$ is $\text{lcm}(2,3,7) = 42$, the order of $(2,1,0,1)$ in $\mathbb{Z}_4 \times \mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_7$ is $\text{lcm}(2,3,1,7) = 42$, the order of $(1,0,3,1)$ in $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_9 \times \mathbb{Z}_7$ is $\text{lcm}(2,1,3,7) = 42$, and the order of $(1,0,1,0,1)$ in $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_7$ is $\text{lcm}(2,1,3,1,7) = 42$.
2. $A_4 \times \mathbb{Z}_2$ is not isomorphic to $S_4$ because $S_4$ has an element of order $4$, for example the $4$-cycle $(1\;2\;3\;4)$, whereas $A_4 \times \mathbb{Z}_2$ does not have an element of order $4$, and having an element of a certain order is a structural property. The elements of $A_4$ have orders $1$, $2$, and $3$ and the elements of $\mathbb{Z}_2$ have orders $1$ and $2$, so taking least common multiples we see that the elements of $A_4 \times \mathbb{Z}_2$ have orders $1$, $2$, $3$, and $6$.

   Alternatively, one could argue that $A_4 \times \mathbb{Z}_2$ has a non-identity element that commutes with every element, namely $(\text{identity}, 1)$, whereas $S_4$ does not, and the property of having a non-identity element that commutes with every element is a structural property.

3. $\operatorname{ker}(\phi)$ is a subgroup of $G$:
   • It is closed under the operation: let $a,b \in \text{ker}(\phi)$. Then $\phi(ab) = \phi(a)\phi(b) = e_{G'}e_{G'} = e_{G'}$, so $ab \in \text{ker}(\phi)$.
   • It contains the identity element: $\phi(e_G) = e_{G'}$, so $e_G \in \text{ker}(\phi)$.
   • It is closed under inverses: let $a \in \text{ker}(\phi)$. Then $\phi(a^{-1}) = \phi(a)^{-1} = (e_{G'})^{-1} = e_{G'}$, so $a^{-1} \in \text{ker}(\phi)$.
4. For this problem it helps to consider the kernel of $\phi$. Because $\ker(\phi)$ is a subgroup of $\mathbb{Z}_p$ its order must divide $p$ by Lagrange's theorem, so its order must be $1$ or $p$ because $p$ is prime. This means that either $\ker(\phi)$ is either the trivial subgroup of $\mathbb{Z}_p$ or the improper subgroup of $\mathbb{Z}_p$ (meaning $\mathbb{Z}_p$ itself.) If $\ker(\phi)$ is trivial then $\phi$ is injective, and if $\ker(\phi)$ is improper (meaning it is equal to the domain of $\phi$) then $\phi$ is trivial.
5. Consider the following two subgroups of $\operatorname{GL}_2(\mathbb{R})$. (You may assume that they are subgroups.) $$\begin{align} H_1 &= \left\{ \begin{pmatrix} x & 0 \\ 0 & x \end{pmatrix} : x \in \mathbb{R}^* \right\}\\ H_2 &= \left\{ \begin{pmatrix} x & 0 \\ 0 & y \end{pmatrix} : x,y \in \mathbb{R}^* \right\}. \end{align}$$
   1. $H_1$ is normal in $\operatorname{GL}_2(\mathbb{R})$: Let $h \in H_1$ and $A \in \operatorname{GL}_2(\mathbb{R})$. We have $h = xI$ for some $x \in \mathbb{R}^*$, where $I$ is the $2 \times 2$ identity matrix. Then $AhA^{-1} = AxIA^{-1} = xAIA^{-1} = xAA^{-1} = xI = h$.
   2. $H_2$ is not normal in $\operatorname{GL}_2(\mathbb{R})$. Consider the matrices $$\begin{align} h &= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \in H_2\\ A &= \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \in \operatorname{GL}_2(\mathbb{R}). \end{align}$$ We have $$ AhA^{-1} = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -2 \\ 0 & -1 \end{pmatrix} \notin H_2.$$ Note: in linear algebra, conjugate matrices are called similar, and what we showed is that (a) every matrix that is similar to a scalar matrix is a scalar matrix (equal to the original, in fact) and (b) not every matrix that is similar to a diagonal matrix (i.e. diagonalizable) is diagonal.
6. Let $G$ be a group and let $H$ be a subgroup of $G$.
   1. Let $a \in G$. If $b_1, b_2 \in aHa^{-1}$, say $b_1 = ah_1a^{-1}$ and $b_2 = ah_2a^{-1}$, then $b_1b_2 = ah_1a^{-1}ah_2a^{-1} = ah_1h_2a^{-1} \in aHa^{-1}$ because $h_1h_2 \in H$. We have $e_G = aa^{-1} = ae_Ga^{-1} \in aHa^{-1}$ because $e_G \in H$. Finally, if $b \in aHa^{-1}$, say $b = aha^{-1}$ where $h \in H$, then $b^{-1} = (aha^{-1})^{-1} = (a^{-1})^{-1}h^{-1}a^{-1} = ah^{-1}a^{-1} \in aHa^{-1}$ because $h^{-1} \in H$.
   2. Let $a \in G$. Note that the subgroup $aHa^{-1}$ of $G$ (called a conjugate subgroup of $H$ in $G$) has the same order as $H$, because the function $f : H \to aHa^{-1}$ defined by $f(h) = aha^{-1}$ is a bijection (it is inverted by the function $x \mapsto a^{-1}xa$.) Then our hypothesis implies that the subgroup $aHa^{-1}$ must be equal to $H$. Therefore $H$ is normal in $G$.