Group Theory Exercises

Let $G$ be a group of order $4$. By Lagrange’s theorem, all elements in $G$ have order $1$, $2$ or $4$. If an element $a$ has order $4$ then $G$ is cyclic,[1] and thus abelian.[2]

Otherwise suppose all non-identity elements have order $2$. Then $G = \{e, a, b, c\}$ and $a^2 = b^2 = c^2 = e$. By the cancellation law in groups, $ab \neq a$, $ab \neq b$ and $ab \neq e$ and therefore $ab = c$. Thus $(ab)^2 = e$, implying $ab = ba$ and the conclusion follows.

Let $|a| = n$ and $|xax^{-1}| = k$. Then $$ \begin{align} (xax^{-1})^n &= \overbrace{(xax^{-1})(xax^{-1})\cdots(xax^{-1})}^{n \text{ times}}\\ &= xa^nx^{-1}\\ &= xex^{-1}\\ &= xx^{-1} = e. \end{align} $$ Thus $k \leq n$. Similarly, $$ \begin{align} e = (xax^{-1})^k &= \overbrace{(xax^{-1})(xax^{-1})\cdots(xax^{-1})}^{k \text{ times}}\\ &= xa^kx^{-1}, \end{align} $$ which implies $a^k = e$. Thus $n \leq k$ and therefore $n = k$.

Let $g \in G$. Then $gN$ is a left coset of $N$ in $G$. Since $[G:N] = 2$, there are only two left cosets of $N$ in $G$, namely $N$ and $gN$. Thus $gN = N$ or $gN = G \setminus N$. In the former case, $g \in N$ and therefore $gN = Ng$. In the latter case, $g \notin N$ and therefore $gN = G \setminus N = Ng$. Thus $gN = Ng$ for all $g \in G$ and therefore $N$ is normal in $G$.

Necessity: Let $G$ be an infinite cyclic group. Then $G$ is isomorphic to $\mathbb{Z}$, which is isomorphic to each of its proper subgroups (namely, $n\mathbb{Z}$ for $n \in \mathbb{N}$).

Sufficiency: Let $G$ be a group isomorphic to each of its proper subgroups. Then if $a \neq e$ is an element of $G$, the subgroup $\langle a \rangle$ generated by $a$ is a proper subgroup of $G$ and therefore $G \cong \langle a \rangle$. Thus $G$ is cyclic. Now suppose $|G| = n \lt \infty$. Then it is isomorphic to $\mathbb{Z}_n$, which is not isomorphic to each of its proper subgroups. Thus $G$ must be infinite.

(1) $\Rightarrow$ (2). If $|G|$ is prime, then $G$ has no proper subgroups by Lagrange’s theorem. Since $|G| \gt 1$, $G \neq \langle e \rangle$.

(2) $\Rightarrow$ (3). If $G \neq \langle e \rangle$ and $G$ has no proper subgroups, then $G$ is cyclic. Since $G$ is finite, it is isomorphic to $\mathbb{Z}_n$ for some $n \in \mathbb{N}$. Since $G$ has no proper subgroups, $n$ must be prime.

(3) $\Rightarrow$ (1). If $G \cong \mathbb{Z}_p$ for some prime $p$, then $|G| = p$.

Since $\alpha$ is an odd permutation, it follows that the equation $$ \alpha = \tau_1 \tau_2 \cdots \tau_{m} $$ where each $\tau_i$ is a transposition is true if and only if $m$ is odd. Let $\beta \in S_n$ be an arbitrary permutation whose parity we do not necessarily know. Then if $$ \beta = \tau'_1 \tau'_2 \cdots \tau'_{k} $$ for some $k \in \mathbb{N}$ (the $\tau_i$'s being transpositions), then $\beta^{-1}$ is also a $k$-cycle, i.e., $$ \beta^{-1} = \tau''_1 \tau''_2 \cdots \tau''_{k}, $$ where again each $\tau''_i$ is a transposition. Thus we have $$ \beta\alpha\beta^{-1} = (\tau'_1 \tau'_2 \cdots \tau'_{k})(\tau_1 \tau_2 \cdots \tau_{m})(\tau''_1 \tau''_2 \cdots \tau''_{k}), $$ and therefore $\beta\alpha\beta^{-1}$ has length $k + m + k = 2k + m$. Since $m$ is odd, there exists a natural number $n$ such that $m = 2n + 1$. Thus $2k + m = 2k + 2n + 1 = 2(k + n) + 1$, which is odd. Therefore $\beta\alpha\beta^{-1}$ is an odd permutation.

An element of $S_7$ of order $12$ must be the product of a disjoint

(1) Let $e$ be the identity element of $H$. Then since $\varphi$ is a homomorphism, we have $$ \begin{aligned} \varphi(xyx^{-1}y^{-1}) &= \varphi(x)\varphi(y)\varphi(x^{-1})\varphi(y^{-1})\\ &= \varphi(x)\varphi(y)\varphi(x)^{-1}\varphi(y)^{-1}. \end{aligned} $$ Since $G$ is abelian, this simplifies to $$ \varphi(x)\varphi(x)^{-1}\varphi(y)\varphi(y)^{-1} = e, $$ and therefore $xyx^{-1}y^{-1} \in \operatorname{Ker} \varphi$ for all $x, y \in G$.

(2) Let