1. (i) Prove or disprove that $\mathbf{Z} \oplus \mathbf{Z}$ is cyclic.

The group is not cyclic. Suppose for contradiction that it is cyclic, generated by $(a, b)$, say. Then $k(a, b)=(1,0)$ for some $k$. But $k a=1$ implies $k \neq 0$, so $k b=0$ means $b=0$. A symmetric argument with $(0,1)$ shows then that $a=0$. But $\langle 0,0\rangle$ is the trivial group.

(ii) Find a generator of the cyclic group $\mathbf{Z}_{10} \oplus \mathbf{Z}_{9}$.

One generator is $(1,1)$. One way to see this is to compute all the powers of $(1,1)$, but a more sophisticated way is to note that the order of 1 in $\mathbf{Z}_{10}$ is 10 , and its order in $\mathbf{Z}_{9}$ is 9 , so the order of $(1,1)$ in $\mathbf{Z}_{10} \times$ $\mathbf{Z}_{9}$ is $\operatorname{lcm}(10,9)=90$. On the other hand, the order of $\mathbf{Z}_{10} \times \mathbf{Z}_{9}$ is $10 \times9=90$. The result follows.

(iii) Prove or disprove that $\mathbf{Z}_{8} \times \mathbf{Z}_{2} \times \mathbf{Z}_{5} \times \mathbf{Z}_{25} \approx \mathbf{Z}_{50} \times \mathbf{Z}_{40}$.

Observe that $\mathbf{Z}_{8} \times \mathbf{Z}_{5} \approx \mathbf{Z}_{40}$, because $\operatorname{lcm}(8,5)=40$. Similarly $\mathbf{Z}_{2} \times$ $\mathbf{Z}_{25} \approx \mathbf{Z}_{50}$. The isomorphism in the question then follows.

2. (i) Find the order of the element $10+\langle 8\rangle$ in the factor group $\mathbf{Z}_{24} /\langle 8\rangle$.

We compute:\begin{gathered}10+\langle 8\rangle=2+\langle 8\rangle \\(10+\langle 8\rangle)+(10+\langle 8\rangle)=20+\langle 8\rangle=4+\langle 8\rangle\end{gathered}It follows that $10+⟨8⟩$ has order 4 in ${\bf Z}_{24}/⟨8⟩$.

(ii) Determine what familiar group $D_{4} / Z\left(D_{4}\right)$ is isomorphic to.

Note that $D_{8}$ has eight elements. The center of $D_{8}$ is $\left\{R_{0}, R_{180}\right\}$ (check this). Thus the number of elements in $D_{4} / Z\left(D_{4}\right)$ is four, and hence it is isomorphic either to $\mathbf{Z}_{4}$ or to $\mathbf{Z}_{2} \times \mathbf{Z}_{2}$. One then checks the orders of elements and observes that there is no element of order four, so the group is not cyclic. Hence it is isomorphic to $\mathbf{Z}_{2} \times \mathbf{Z}_{2}$.

3. Suppose $H$ and $K$ are both normal subgroups of a group $G$, and they have the property that $H \cap K=\{e\}$.

(i) Show that elements from $H$ commute with elements from $K$. (Hint: Do this by showing that $h k h^{-1} k^{-1}=e$ for any $h \in H$ and $k \in K$.)

Looking at the product in the hint as $\left(h k h^{-1}\right) k^{-1}$, we see that the term in the parentheses lies in $K$, because $K$ is normal. Thus its product with $k^{-1}$ also lies in $K$. On the other hand, parenthesizing as $h\left(k h^{-1} k^{-1}\right)$, the normality of $H$ implies that this element lies in $H$. Since $H$ and $K$ intersect trivially, it follows that $h k h^{-1} k^{-1}=e$, so that elements of $H$ commute with elements of $K$.

(ii) Suppose also that $G=H K$. Prove that $G \approx H \oplus K$. (Hint: Construct the obvious map $\varphi: H \oplus K \rightarrow G$ and show it is an automorphism.)

We consider the map $\varphi(h, k)=h k$ and show that it is a one-to-one and onto homomorphism.

To see that it is one-to-one, suppose that $\varphi\left(h_1,k_1\right)=\varphi\left(h_2,k_2\right)$. Then $h_1k_1=h_2k_2$. Doing some algebra gives $h_{2}^{-1} h_{1}=k_{2} k_{1}^{-1}$. But the left-hand side is in $H$ while the right-hand side is in $K$. Since $H$ and $K$ intersect trivially, it follows that $h_2^{-1}h_2$ and $k_2k_1^{-1}$ are both trivial, and hence that $h_1=h_2$ and $k_1=k_2$.

The fact that it is onto follows follows form the fact that $G=H K$.

To see that it is a homomorphism, consider\begin{aligned} \varphi\left(\left(h_{1}, k_{1}\right) \cdot\left(h_{2}, k_{2}\right)\right) &=\varphi\left(h_{1} h_{2}, k_{1} k_{2}\right)=\left(h_{1} h_{2}\right)\left(k_{1} k_{2}\right)=h_{1}\left(h_{2} k_{1}\right) k_{2}=h_{1}\left(k_{1} h_{2}\right) k_{2} \\ &=\left(h_{1} k_{1}\right)\left(h_{2} k_{2}\right)=\varphi\left(h_{1}, k_{1}\right) \cdot \varphi\left(h_{2}, k_{2}\right) \end{aligned}Note that we have used the commutivity of $H$ with $K$ to change $h_{2} k_{1}$ into $k_{1} h_{2}$.

4. (i) Let $\mathbf{Z}[x]$ denote the group of all polynomials with integer coefficients under addition, and let $f^{\prime}$ denote the derivative of $f \in \mathbf{Z}[x]$. Show that the map $\theta: f \mapsto f^{\prime}$ is a homomorphism from $\mathbf{Z}[x]$ to itself.

We observe that$$\theta(f+g)=(f+g)^{\prime}=f^{\prime}+g^{\prime}=\theta(f)+\theta(g)$$(ii) Use the first isomorphism theorem to prove the second isomorphism theorem: If $K$ is a subgroup of $G$ and $N$ is a normal subgroup of $G$, prove that $K /(K \cap N)$ is isomorphic to $K N / N$.

We define a map $\varphi: K \rightarrow K N / N$ by $\varphi(k)=k N(=k e N)$.

This is a homomorphism: $\varphi\left(k_{1} k_{2}\right)=\left(k_{1} k_{2}\right) N=\left(k_{1} N\right)\left(k_{2} N\right)=\varphi\left(k_{1}\right) \varphi\left(k_{2}\right)$.

It is onto: Given $k n N \in K N / N$, we note that $k n N=k N=\varphi(k)$.

The kernel is $K \cap N: \varphi(k)=k N=N \quad \Leftrightarrow \quad k \in N \quad \Leftrightarrow \quad k \in K \cap N$.

5. (i) Find all abelian groups (up to isomorphism) of order 180. (Be sure that your list contains no duplicates.)

Observe that 180 factors as $2 \times 2 \times 3 \times 3 \times 5$. To avoid repetition, we only combine common factors. Thus we have

$\mathbf{Z}_{2} \times \mathbf{Z}_{2} \times \mathbf{Z}_{3} \times \mathbf{Z}_{3} \times \mathbf{Z}_{5}, \quad \mathbf{Z}_{4} \times \mathbf{Z}_{3} \times \mathbf{Z}_{3} \times \mathbf{Z}_{5}, \quad \mathbf{Z}_{2} \times \mathbf{Z}_{2} \times \mathbf{Z}_{9} \times \mathbf{Z}_{5},\quad \mathbf{Z}_{4} \times \mathbf{Z}_{9} \times \mathbf{Z}_{5}$

Note that this last one is isomorphic to $\mathbf{Z}_{180}$.

(ii) Suppose $G_{1}$ and $G_{2}$ are groups. Show that $G_{1} \times\{e\}$ is a normal subgroup of $G_{1} \times G_{2}$, and show that $\left(G_{1} \times G_{2}\right) /\left(G_{1} \times\{e\}\right) \approx G_{2}$.

Let $\left(g_{1}, g_{2}\right)$ be an arbitrary element of $G_{1} \times G_{2}$, and let $(h, e)$ be an element of $G_{1} \times\{e\}$. Then we have$$\left(g_{1}, g_{2}\right)(h, e)\left(g_{1}^{-1}, g_{2}^{-1}\right)=\left(g_{1} h g_{1}^{-1}, g_{2} e g_{2}^{-1}\right)=\left(g_{1} h g_{1}^{-1}, e\right) \in G_{1} \times\{e\}$$

so that $G_{1} \times\{e\}$ is normal in $G_{1} \times G_{2}$.

Define $\varphi: G_{1} \times G_{2} \rightarrow G_{2}$ by $\varphi\left(g_{1}, g_{2}\right)=g_{2}$. Clearly $\varphi$ is a homomorphism mapping onto $G_{2}$. Also $\varphi\left(g_{1}, g_{2}\right)=e$ if and only if $g_{2}=e$, so that $\operatorname{ker} \varphi=\left\{\left(g_{1}, e\right)\right\}=G_{1} \times\{e\}$. he result now follows from the first isomorphism theorem.

6. Let $G$ be a finite group and $H$ a normal subgroup of $G$.

(i) Prove that the order of $g H$ in $G / H$ divides $|g|$ in $G$.

By the 'normal subgroups are kernels' theorem, there is a homomorphism $\varphi: G \rightarrow G$ such that $\operatorname{ker} \varphi=H$ and $\operatorname{im} \varphi \approx G / H$. Hence the order of $g H$ in $G / H$ is the same as the order of $\varphi(g)$, which divides $|g|$ by theorem 10.1(3).

(ii) Prove that if $G / H$ has an element of order $n$, then $G$ has an element of order $n$.

We use part (i). Suppose $g H$ has order $n$ in $G / H$. Then the order of $g$ is divisible by $n$, say $|g|=n k$. But then $g^{k}$ has order $n$ in $G$.

7. (i) Show that any group homomorphism $\varphi: G \rightarrow \bar{G}$ where $|G|$ is prime must be either the trivial homomorphism or a one-to-one map.

Because $|G|$ is prime, we know from Lagrange's theorem that the only subgroups of $G$ are the trivial group and $G$ itself. Thus the kernel of $\varphi$ is either trivial, in which case $\varphi$ is one-to-one, or $G$ itself, in which case $\varphi$ is trivial.

(ii) Recall that $\operatorname{Aut}(G)$ is the group of automorphisms of $G$, while $\operatorname{Inn}(G)$ is the subgroup of all inner automorphisms of $G$; i.e., an element $T_{g} \in \operatorname{Inn}(G)$ has the form $T_{g}(h)=g h g^{-1}$. Show that $\operatorname{Inn}(G)$ is a normal subgroup of $\operatorname{Aut}(G)$. (The factor group $\operatorname{Aut}(G) / \operatorname{Inn}(G)$ is called the

Let $\varphi$ be an arbitrary automorphism of $G$, and $f i x$ an inner automorphism $T_{g}$. Then for any group element $h \in G$ we have\begin{gathered}\left(\varphi \circ T_{g}\right)(h)=\varphi\left(g h g^{-1}\right)=\varphi(g) \varphi(h) \varphi\left(g^{-1}\right)=\varphi(g) \varphi(h) \varphi(g)^{-1} \\=T_{\varphi(g)} \varphi(h)=\left(T_{\varphi(g)} \circ \varphi\right)(h)\end{gathered}Thus $\varphi \circ T_{g}=T_{\varphi(g)} \circ \varphi$, so that $\varphi \operatorname{Inn}=\operatorname{Inn} \varphi$.