Artin, Algebra, Chapter 2 Exercises

Problem 1.1. Let $S$ be a set. Prove that the law of composition defined by $ab = a$ for all $a$ and $b$ in $S$ is associative. For which sets does this law have an identity?

Solution. Given $a,b,c \in S$, one has $$(ab)c = ac = a = ab = a(bc),$$ so this law of composition on $S$ is associative. Next, I claim that this law has an identity element if and only if $S$ is a singleton set. Indeed, if $S = \{s\}$, then $ss = s$, so $s$ is the identity element. Conversely, suppose $S$ contain an identity element $e$. Then, for any $a \in S$, one has $ae = a = ea$. But the law of composition implies that $ea = e$, hence $a = e$ and $S = \{e\}$.

Problem 1.3. Let $\mathbb{N}$ denote the set $\{1,2,3\}$ of natural numbers, and let $s: \mathbb{N} \to \mathbb{N}$ be the shift map defined by $s(n) = n + 1$. Prove that $s$ has no right inverse, but that it has infinitely many left inverses.

Solution. Suppose, for the sake of contradiction, that there exists a right inverse $g: \mathbb{N} \to \mathbb{N}$ for $s$. Then $(s \circ g)(n) = n$ for every $n \in \mathbb{N}$. In particular, one has $(s \circ g)(1) = 1$, from which it follows that $$1 = (s \circ g)(1) = s(g(1)) = g(1) + 1,$$ in which case $g(1) = 0 \not \in \mathbb{N}$, which is a contradiction. So no such right inverse $g$ exists.

Now, fix $m \in \mathbb{N}$, and define $f_m: \mathbb{N} \to \mathbb{N}$ by $$f_m (n) = \begin{cases} m - 1 & \text{ if $n \neq 1$,} \\ m & \text { otherwise.} \end{cases}$$ Then, for any $n \in \mathbb{N}$, one has $$(f_m \circ s)(n) = f_m (s(n)) = f_m (n+1) = (n+1)-1 = n,$$ so $f_m$ is a left inverse of $s$. As $m$ can be chosen arbitrarily from $\mathbb{N}$, we have in fact defined a countable family of left inverses.

Problem 2.1. Make a multiplication table for the symmetric group $S_3$.

Solution. The standard presentation for the symmetric group $S_3$, given in (2.2.7) in Artin, is $$S_3 = \left \langle x, y \mid x^3 = 1 = y^2, \; yx = x^2 y \right \rangle = \left\{1, x, x^2, y, xy, x^2 y \right\},$$ where $x = (123)$ and $y = (12)$. Using these defining relations, we compute the multiplication table for $S_3$.

$\circ$ $1$ $x$ $x^2$ $y$ $xy$ $x^2 y$
$1$ $1$ $x$ $x^2$ $y$ $xy$ $x^2 y$
$x$ $x$ $x^2$ $1$ $xy$ $x^2y$ $y$
$x^2$ $x^2$ $1$ $x$ $x^2 y$ $y$ $x y$
$y$ $y$ $x^2 y$ $xy$ $1$ $x^2$ $x$
$xy$ $xy$ $y$ $x^2 y$ $x$ $1$ $x^2$
$x^2 y$ $x^2 y$ $xy$ $y$ $x^2$ $x$ $1$

Problem 4.3. Let $a$ and $b$ be elements of a group $G$. Prove that $ab$ and $ba$ have the same order.

Solution. We begin with a lemma.

Lemma. For any $n \in \mathbb{N}$, $$(ab)^n = a(ba)^{n-1} b.$$ Proof. We proceed by induction on $n$. When $n = 1$, we find that $(ab)^1 = ab = aeb = a(ba)^{1-1} b$. Suppose inductively that $(ab)^k = a(ba)^{k-1} b$ for some fixed $k \geq 1$. It then follows that $$(ab)^{k+1} = (ab)^k (ab) = (a(ba)^{k-1} b)(ab) = a((ba)^{k-1} (ba))b = a(ba)^k b,$$ which closes the induction. $\blacksquare$

Now, for any $n \in \mathbb{N}$, we have $(ab)^n = e$ if and only if $a(ba)^{n-1} b = e$, which is true if and only if $(ba)^{n-1} = a^{-1} b^{-1} = (ba)^{-1}$, which is true if and only if $(ba)^n = e$. So $ab$ has finite order if and only if $ba$ has finite order, hence $ab$ has infinite order if and only if $ba$ has infinite order. Furthermore, if $ab$ and $ba$ have finite order, then the forward direction implies that the order of $ab$ divides the order of $ba$ and the backward direction that the order of $ba$ divides the order of $ab$. As $|ab|$ and $|ba|$ are positive, we then conclude that $|ab| = |ba|$.

Problem 5.1. Let $\varphi: G \to G'$ be a surjective homomorphim. Prove that if $G$ is cyclic, then $G'$ is cyclic, and if $G$ is abelian, then $G'$ is abelian.

Solution. Suppose that $G$ is cyclic. So there exists $a \in G$ such that $G = \langle a \rangle$. Fix $y' \in G'$. As $\varphi$ is surjective, there exists $x' \in G$ such that $\varphi(x') = y'$. As $G$ is cyclic, there exists $m \in \mathbb{Z}$ such that $x' = a^m$. Then $$y' = \varphi(x') = \varphi(a^m) = \varphi(a)^m,$$ so $y' \in \langle \varphi(a) \rangle$, hence $\langle \varphi(a) \rangle \supseteq G'$. As $\langle \varphi(a) \rangle \subseteq G'$ by definition, we conclude that $\langle \varphi(a) \rangle = G'$, so $G'$ is cyclic.

Next, let $x',y' \in G$. By surjectivity of $\varphi$, there exist $x,y \in G$ such that $\varphi(x) = x'$ and $\varphi(y) = y'$. As $G$ is abelian, we find that $$x' y' = \varphi(x) \varphi(y) = \varphi(xy) = \varphi(yx) = \varphi(y) \varphi(x) = y'x',$$ so $G'$ is abelian, as required.

Problem 5.2. Prove that the intersection of $K \cap H$ of subgroups of a group $G$ is a subgroup of $H$, and that if $K$ is a normal subgroup of $G$, then $K \cap H$ is a normal subgroup of $H$.

Proof. We have $K \cap H \subseteq G$ by definition $K,H \subseteq G$. Let $x,y \in K \cap H$. Then $x,y \in K$ and $x,y \in H$. As $K$ and $H$ are closed under composition, we have $xy \in K$ and $xy \in H$, so $xy \in K \cap H$. As $K$ and $H$ are subgroups, they contain the identity element of $G$, so $e \in K \cap H$. Finally, given $x \in K \cap H \subseteq G$, there exists an inverse $x^{-1} \in G$. As $K$ and $H$ are closed under inversion, we have $x^{-1} \in K$ and $x^{-1} \in H$, so $x^{-1} \in K \cap H$. So $K \cap H$ is a subgroup of $G$.

Suppose now that $K$ is a normal subgroup of $G$. Let $a \in K \cap H$ and $b \in H$. Then $a \in H$, $b \in H$, and $b^{-1} \in H$, so $bab^{-1} \in H$. Furthermore, one has $a \in K,$ so because $K$ is a normal subgroup of $G \ni b$, we have $bab^{-1} \in K$. So $bab^{-1} \in K \cap H$, so $K \cap H$ is a normal subgroup of $H$.

Problem 6.1. Let $G'$ be the group of real matrices of the form $\begin{bmatrix} 1 & x \\ & 1 \end{bmatrix}$. Is the map $\mathbb{R}^{+} \to G'$ that sends $x$ to this matrix an isomorphism?

Solution. Define $\varphi: \mathbb{R}^{+} \to G'$ by $\varphi(x) = \begin{bmatrix} 1 & x \\ 0 & 1 \end{bmatrix}$. I claim that $\varphi$ is an isomorphism. First, if $\varphi(x) = \varphi(y)$ for $x,y \in \mathbb{R}$, then $\begin{bmatrix} 1 & x \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix}$. Equating entries then yields $x = y$, so $\varphi$ is injective. Second, given $\begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} \in G'$, we have $\varphi(a) = \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix}$, so $\varphi$ is surjective and hence bijective. Finally, we will show that $\varphi$ is a homomorphism. For any $g,h \in \mathbb{R}$, we find that $$\varphi(g+h) = \begin{bmatrix} 1 & g +h \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & g \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & h \\ 0 & 1 \end{bmatrix} = \varphi(g) \varphi(h),$$ so $\varphi$ is a homomorphism and hence an isomorphism, hence $\mathbb{R}^{+} \cong G'$, as required.

Problem 6.4. Prove that in a group, the products $ab$ and $ba$ are conjugate elements.

Solution. Let $a,b \in G$. We have $$ab = e(ab) = \left(b^{-1} b\right)(ab) = b^{-1} (ba)b,$$ where $b = \left(b^{-1}\right)^{-1}$ , so $ab$ and $ba$ are conjugate elements.

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