HW1

(1) Let $\mathbb{Q}$ be the additive group of rational numbers.

• (a) Show that it is not a finitely generated group.
• (b) Show that it is not a free abelian group.

\begin{proof} (a) Otherwise, assume $\mathbb{Q}=⟨x_i:i\in I⟩$ where $x_i\in \mathbb{Q}$ and $|I|<\infty$. Suppose that $x_i=p_i/q_i$ with $(p_i,q_i)=1$. Define $M:=\{p_i:i\in I\}\cup \{q_i:i\in I\}$ and define $P:=\{p:p\text{mbox}isaprime,p\mid m\text{mbox}forsomem\in M\}$. Since $|I|<\infty$, we have that $|P|<\infty$. Then there is a prime $q$ such that $q\notin P$ and so $1/q\notin ⟨x_i:i\in I⟩$. It is a contradiction because $\mathbb{Q}=⟨x_i:i\in I⟩$. Therefore, $\mathbb{Q}$ is not a finitely generated group.

(b) Assume that $\mathbb{Q}$ is a free abelian group, then $\mathbb{Q}$ has a basis $X$. We claim that $|X|=1$. Otherwise, take $x_1,x_2\in X$ with $x_1=p_1/q_1$, $x_2=p_2/q_2$. Then $(p_2{q}_1)\cdot x_1+(-p_1{q}_2)\cdot x_2=0$ yields that $x_1{x}_2=0$, which is impossible. So $|X|=1$ and $\mathbb{Q}$ is finitely generated. It contradicts with (a). Therefore, $\mathbb{Q}$ is not a free abelian group. \end{proof}

(2) Recall $A_n◃S_n$ is the normal subgroup called the alternating groups ($A_n$ consists of permutations that can be written as a product of even numbers of transpositions). Show that any finite group $G$ is isomorphic to a subgroup of $A_n$ for some $n$.

\begin{proof} Let $n:=|G|+2$. There exists a group action $f$ of $G$ on itself, where $f:G\times G\to G,(g,h)\mapsto gh$. Thus $G$ is isomorphic to a subgroup of $S_{n-2}$. Define $\varphi :S_{n-2}\to A_n$ as

$$\varphi (\tau )=\{\begin{array}{ll}\tau & \tau \in A_{n-2}\\ \tau \circ (n-1n)& \tau \notin A_{n-2}\end{array},\text{mbox}forall\tau \in S_{n-2}.$$

It is easy to verify that $\varphi$ is a group homomorphism and $\varphi$ is injective. So $S_{n-2}$ is isomorphic to a subgroup of $A_n$, and then $G$ is isomorphic to a subgroup of $A_n$. \end{proof}

(3) Let $G$ be a finite group of even order. Then $G$ contains an element $a\ne e_G$ such that $a^2=e_G$.

\begin{proof} Assume that $g^2\ne e_G$ for all non-identity $g\in G$, then $g\ne g^{-1}$ for all non-identity $g\in G$. So there exists $g_1,\cdots ,g_n\in G$ such that $G=\{e_G,g_1,g_1^{-1},\cdots ,g_n,g_n^{-1}\}$. It contradicts with $|G|$ even. Therefore, $G$ contains an element $a\ne e_G$ such that $a^2=e_G$. \end{proof}

(4) Let $M=\mathbb{Z}\cdot e_1\oplus \mathbb{Z}\cdot e_2$. Let $M^{\prime }=⟨2e_1+20e_2,2e_1+30e_2⟩$ be a subgroup. Find a new basis $\left(f_1,f_2\right)$ of $M$ so that $M^{\prime }$ can be written as $⟨a_1{f}_1,a_2{f}_2⟩$ with $a_1\mid a_2$.

Solution. Note that

$$\begin{array}{rl}\left[\begin{array}{ll}2& 20\\ 2& 30\end{array}\right]\left[\begin{array}{l}{e}_1\\ e_2\end{array}\right]& =\left[\begin{array}{ll}1& \\ 1& 1\end{array}\right]\left[\begin{array}{ll}2& 20\\ & 10\end{array}\right]\left[\begin{array}{l}{e}_1\\ e_2\end{array}\right]\\ & =\left[\begin{array}{ll}1& \\ 1& 1\end{array}\right]\left[\begin{array}{ll}2& \\ & 10\end{array}\right]\left[\begin{array}{cc}1& 10\\ & 1\end{array}\right]\left[\begin{array}{l}{e}_1\\ e_2\end{array}\right]\\ & =\left[\begin{array}{ll}1& \\ 1& 1\end{array}\right]\left[\begin{array}{ll}2& \\ & 10\end{array}\right]\left[\begin{array}{c}{e}_1+10e_2\\ e_2\end{array}\right].\end{array}$$

So $M^{\prime }=⟨2f_1,10f_2⟩$ where $f_1=e_1+10e_2$ and $f_2=e_2$. \end{proof}

(5) Let $M=\mathbb{Z}\cdot e_1\oplus \mathbb{Z}\cdot e_2$ be a free abelian group of rank $2$ . Let $M^{\prime }=⟨20e_1,26e_2⟩$ be a subgroup. Find a new basis $\left(f_1,f_2\right)$ of $M$ so that $M^{\prime }$ can be written as $⟨a_1{f}_1,a_2{f}_2⟩$ with $a_1\mid a_2$.

Solution. Since

$$\left[\begin{array}{cc}1& 1\\ 39& 40\end{array}\right]\left[\begin{array}{cc}20& \\ & 26\end{array}\right]\left[\begin{array}{c}{e}_1\\ e_2\end{array}\right]=\left[\begin{array}{cc}2& \\ & 260\end{array}\right]\left[\begin{array}{c}10e_1+13e_2\\ 3e_1+4e_2\end{array}\right]$$

and $\det \left[\begin{array}{cc}10& 13\\ 3& 4\end{array}\right]=1$, then $M^{\prime }=⟨2f_1,260f_2⟩$ where $f_1=10e_1+13e_2$ and $f_2=3e_1+4e_2$. \end{proof}

(6) Let $G$ be a finite abelian group which is not cyclic. Show that it contains some subgroup of the form $\mathbb{Z}/p\mathbb{Z}\oplus \mathbb{Z}/p\mathbb{Z}$ for some prime number $p$.

\begin{proof} By fundamental theorem of finite abelian groups, $G$ is isomorphic to ${\oplus }_{i=1}^k\mathbb{Z}/p_i^{s_i}\mathbb{Z}$ for some prime $p_i$ and $s_i\in {\mathbb{N}}_{+}$. If $p_1,\cdots ,p_k$ are distinct primes, then $⟨(a_1,\cdots ,a_k)⟩=G$ where $⟨a_i⟩=\mathbb{Z}/p^{s_i}\mathbb{Z}$ by Chinese remainder theorem. So $G$ is cyclic, which is impossible. Therefore, there exists $p_i$, $p_j$ such that $p_i=p_j$, $i\ne j$.

Assume that $p_i=p_j=p$ and $s_i=k_1$, $s_j=k_2$. Then $G$ has a subgroup $\mathbb{Z}/p^{k_1}\mathbb{Z}\oplus \mathbb{Z}/p^{k_2}\mathbb{Z}$. Suppose that $\mathbb{Z}/p^{k_1}\mathbb{Z}=⟨\alpha ⟩$ and $\mathbb{Z}/p^{k_2}\mathbb{Z}=⟨\beta ⟩$. Then $⟨(p^{k_1-1})\alpha ⟩\oplus ⟨(p^{k_2-1})\beta ⟩$ is a subgroup of $G$ and it is isomorphic to $\mathbb{Z}/p\mathbb{Z}\oplus \mathbb{Z}/p\mathbb{Z}$. \end{proof}

(7) Find all the subgroups of $\mathbb{Z}/p^2\mathbb{Z}\oplus \mathbb{Z}/p^3\mathbb{Z}$ of order $p^2$. (Not just isomorphic ones; you need to find ALL possible such subgroups).

Solution. Assume that $\mathbb{Z}/p^2\mathbb{Z}\oplus \mathbb{Z}/p^3\mathbb{Z}=⟨a⟩\oplus ⟨b⟩$.

All subgroups isomorphic to $\mathbb{Z}/p^2\mathbb{Z}$ are

• $⟨(a,0)⟩$,
• $⟨(ka,pb)⟩$ with $k\in \{0,\cdots ,p^2-1\}$, and
• $⟨(ka,p^{2}b)⟩$ with $k\in \{1,\cdots ,p-1\}$.

All subgroups isomorphic to $\mathbb{Z}/p\mathbb{Z}\oplus \mathbb{Z}/p\mathbb{Z}$ are $⟨pa⟩\oplus ⟨p^{2}b⟩$. \end{proof}