Math 120A HW 10 Solutions
- Let $G$ be a group and let $H$ be a normal subgroup of $G$.
- If $G$ is abelian, then for all elements $aH$ and $bH$ of the factor group $G/H$ we have $(aH)(bH) = (ab)H = (ba)H = (bH)(aH)$.
- The symmetric group $S_3$ is not abelian, but the normal subgroup $A_3$ and the factor group $S_3/A_3$ are both abelian (they are isomorphic to $\mathbb{Z}_3$ and $\mathbb{Z}_2$ respectively.)
- We have $(S_3/A_3) \times A_3 \cong \mathbb{Z}_2 \times \mathbb{Z}_3$, which is abelian, so it cannot be isomorphic to $S_3$, which is not abelian.
- We have $a^mH = (aH)^m$, which is the identity element $H$ of $G/H$ by Lagrange's theorem (the order of every element of $G/H$ divides the order of $G/H$ itself, which is $m$.) From $a^mH = H$ it follows that $a^m \in H$.
- We can define a function $\phi : S_n/A_n \to \mathbb{Z}_2$ by
$$\phi(\sigma A_n) = \begin{cases}
0 & \text{if $\sigma$ is even}\\
1 & \text{if $\sigma$ is odd.}
\end{cases}$$
This gives a well-defined function: if $\sigma A_n = \sigma' A_n$ then
either $\sigma$ and $\sigma'$ are both even (in the case $\sigma A_n = \sigma'
A_n = A_n$) or $\sigma$ and $\sigma'$ are both odd (in the case $\sigma A_n =
\sigma' A_n = S_n - A_n$) and in both cases we have $\phi(\sigma A_n)
= \phi(\sigma' A_n)$.
Clearly $\phi$ is a bijection.
To see that $\phi$ is a homomorphism, we check four cases: even times
even is even and $0+0 = 0$; even times odd is odd and $0+1=1$; odd times
even is odd and $1 + 0 = 1$; odd times odd is even and $1+1=0$.
-
Let $H$ be the cyclic subgroup of $\mathbb{Z} \times \mathbb{Z}$ generated by $(1,1)$.
-
$(i,j) + H = (k,l) + H \iff (i,j) - (k,l) \in H \iff i-k = j-l$.
-
Consider an arbitrary element $(i,j) + H \in (\mathbb{Z} \times \mathbb{Z}) / H$.
We have $(i,j) + H = (0,j-i) + H = (j-i)((0,1)+H) \in \langle (0,1)+H \rangle$.
-
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 $n=0$.
-
Define a function $\phi : (G_1 \times G_2)/(H_1\times H_2) \to
(G_1/H_1)\times (G_2/H_2)$ by $\phi( (a_1,a_2)(H_1\times H_2)) =
(a_1H_1,a_2H_2)$.
This is well-defined and injective:
$$\begin{align*}
(a_1,a_2)(H_1\times H_2) = (a_1',a_2')(H_1\times H_2)
&\iff a_1H_1 \times a_2H_2 = a_1'H_1 \times a_2'H_2\\
& \iff a_1H_1 = a_1'H_1 \text{ and } a_2H_2 = a_2'H_2 \\
& \iff (a_1H_1,a_2H_2) = (a_1'H_1,a_2'H_2).
\end{align*}$$
Clearly $\phi$ is surjective.
It is a homomorphism:
$$\begin{align*}
\phi(( (a_1,a_2)(H_1\times H_2))( (b_1,b_2)(H_1 \times H_2)))
&= \phi(((a_1,a_2)(b_1,b_2))(H_1 \times H_2))\\
&= \phi((a_1b_1,a_2b_2)(H_1\times H_2))\\
&= (a_1b_1H_1,a_2b_2H_2)\\
&= ((a_1H_1)(b_1H_1),(a_2H_2)(b_2H_2))\\
&= (a_1H_1,a_2H_2)(b_1H_1,b_2H_2)\\
&= \phi((a_1,a_2)(H_1\times H_2))\phi((b_1,b_2)(H_1\times H_2)).
\end{align*}$$
- Compute the following factor groups. (To "compute" a factor group means to find a more familiar group that is isomorphic to it. More precisely: in this case the factor groups are finite abelian groups, so you can classify them according to the fundamental theorem of finite abelian groups.)
- First we compute $\langle (1,2) \rangle =
\{(0,0),(1,2),(2,0),(3,2)\}$. The factor group $(\mathbb{Z}_4 \times
\mathbb{Z}_4) / \langle (1,2) \rangle$ has order $16/4=4$. The element
$(1,1) + \langle (1,2)\rangle$ of the factor group has order $4$ because $(1,1),(2,2),(3,3)
\notin \langle (1,2)\rangle$. Therefore the factor group is cyclic and
is isomorphic to $\mathbb{Z}_4$.
- First we compute $\langle (2,2)\rangle = \{(0,0),(2,2)\}$.
The factor group $(\mathbb{Z}_4 \times \mathbb{Z}_4) / \langle (2,2)
\rangle$ has order $16/2 = 8$, so because it is abelian the fundamental
theorem of abelian groups says it must be isomorphic to one of the
possibilities $\mathbb{Z}_8$, $\mathbb{Z}_4 \times \mathbb{Z}_2$,
or $\mathbb{Z}_2\times \mathbb{Z}_2 \times \mathbb{Z}_2$.
The order of an element of the group $\mathbb{Z}_4 \times \mathbb{Z}_4$ is the $\text{lcm}$ of the orders of two elements of the group $\mathbb{Z}_4$, so it is $1$, $2$, or $4$.
In general
the order of a coset $aH$ in $G/H$ divides the order of $a$ in $G$. So
in this case the factor group cannot have an element of order $8$,
and therefore cannot be isomorphic to $\mathbb{Z}_8$. The element
$(1,0) +\langle (2,2)\rangle$ of the factor group has order $4$ because
$(1,0),(2,0),(3,0) \notin \langle (2,2)\rangle$ but $4(1,0) = (0,0)
\in \langle (2,2)\rangle$. Therefore the factor group is isomorphic to
$\mathbb{Z}_4 \times \mathbb{Z}_2$ (not $\mathbb{Z}_2 \times \mathbb{Z}_2
\times \mathbb{Z}_2$, which has no element of order $4$.)
- We have $\{(0,0),(0,2),(2,0),(2,2)\} = \langle 2 \rangle \times \langle 2 \rangle$, so $(\mathbb{Z}_4 \times \mathbb{Z}_4) / \{(0,0),(0,2),(2,0),(2,2)\} = (\mathbb{Z}_4\times \mathbb{Z}_4)/(\langle 2 \rangle \times \langle 2 \rangle) \cong (\mathbb{Z}_4 / \langle 2 \rangle) \times (\mathbb{Z}_4 / \langle 2 \rangle) \cong \mathbb{Z}_2 \times \mathbb{Z}_2$.