I Groups

I.1 Semigroups, Monoids, and Groups

I.1.2

Exercise. Let $(G, +)$ be a group, $S$ a non-empty set, and $M(S, G)$ the set of all functions $f: S\rightarrow G$. Define addition in $M(S, G)$ as follows: $(f+g):S \rightarrow G$ is given by $s\mapsto f(s) + g(s) \in G$. Prove that $M(S, G)$ is a group, which is abelian if $G$ is.

Proof. First, we consider $f, g, h \in M(S, G)$. We see that $(f+g) +h$ is defined by $s \mapsto (f(s)+g(s)) + h(s)$. Since $+$ in $G$ is associative, we know that this is the same as saying $s \mapsto f(s) +(g(s)+h(s))$. So $(f+g) +h = f+(g+h)$, that is, $+$ in $M(S, G)$ is associative. Next, we consider the identity function $i: S \rightarrow G$ defined by $s \mapsto e_G$, and note that $f+i$ is given by $s\mapsto f(s) + i(s)$, where we can see that $f(s) + i(s) = f(s) + e_{G}= f(s)$, and $f(s) = e_{G} + f(s)=i(s) + f(s)$. So we have that $f= f+i=i+f$, that is, $i$ is the identity element of $M(S, G)$. Now let $-f \in M(S, G)$ be defined by $s \mapsto -f(s)$, which we know exists in $G$. Then $f + (-f)$ is defined by $s\mapsto f(s) + - f(s) = e_{G}= i(s)$, and similarly, $(-f) + f$ is defined by $s\mapsto -f(s) + f(s) = e_{G}= i(s)$. So $f+(-f) = (-f) + f = i$, that is, every element of $M(S, G)$ has an inverse.

Lastly, we note that if $G$ is abelian, $f+g$ is given by $s \mapsto f(s) + g(s) = g(s) + f(s)$, that is, $f+g = g+f$ for all $f, g \in M(S,G)$. Thus, if $G$ is abelian, so is $M(S, G)$.

I.2 Homomorphisms and Subgroups

I.2.1

Exercise. If $f: G\rightarrow H$ is a homomorphism of groups, then $f(e_{G})= e_H$ and $f(a^{-1})= f(a)^{-1}$ for all $a \in G$. Show by example that the first conclusion may be false if $G, H$ are monoids that are not groups.

Proof. Let $g \in G$. Then $f(g)=f(ge_{G})=f(g)f(e_{G})$. Since $f(g) \in H$, we know $f(g)^{-1} \in H$, and we may cancel on the left to obtain that $f(e_G)=e_H$. Now $f(e_{G})=e_{H}$ implies that $f(aa^{-1})=e_{H}$ for all $a \in G$. Since $f$ is a homomorphism, we have that $f(aa^{-1})=f(a)f(a^{-1}) = e_H$, and conclude that $f(a^{-1})$ is a right inverse of $f(a)$. By considering $f(a^{-1}a)$ in an analogous argument, we conclude that $f(a^{-1})$ is also a left inverse of $f(a)$. Hence, $f(a^{-1}) = f(a)^{-1}$.

Example. TODO

I.6 Symmetric, Alternating, and Dihedral Groups

I.6.1

Exercise. Find four different subgroups of $S_{4}$ that are isomorphic to $S_3$ and nine different subgroups that are isomorphic to $\mathbb{Z}_2$.

Answer. For subgroups of $S_4$ isomorphic to $S_{3}$, consider the subgroups that consist of all permutations that fix one point (It can be easily checked that these are indeed subgroups.) There are four such subgroups (each corresponding to fixing 1, 2, 3, or 4). Now each subgroups consists of all the permutations on 3 elements, so they are isomorphic to $S_3$.

For subgroups isomorphic to $\mathbb{Z}_2$, we only need to look for elements of order 2. Luckily, there are 9 such elements—6 transpositions and 3 double transpositions. So the subgroups generated by these are 9 different subgroups of $S_4$ isomorphic to $\mathbb{Z}_2$.

I.6.6

Exercise. Prove that $A_n$ is the only subgroup of $S_n$ with index 2. [Hint: Use the fact that the set of $3$-cycles of $S_n$ generates $A_n$.]

Proof. Suppose $H$ is a subgroup of $S_n$ with index 2, and note that $H$ is normal. Now suppose there exists a $3$-cycle, $\sigma$, that is not in $H$. Then $\sigma H$ is the other only other coset of $H$ that is not $H$ itself. Since $H$ is a subgroup, we also know that $\sigma^{-1}\notin H$. So $\sigma^{-1}H$ is also a coset that is not $H$, that is, $\sigma H = \sigma^{-1}H$. Then multiplying by $\sigma$ on both sides, we see that $\sigma^{2 }H= H$. But $\sigma$ is a $3$-cycle, so $\sigma^{2} = \sigma^{-1}$, and we are left with $\sigma^{-1}H = H$, a contradiction. Hence, any subgroup with index 2 must contain all $3$-cycles.

We finish the proof by applying the hint given in the problem (Lemma I.6.11 in the text), and conclude that $H$ must be $A_{4}$.

I.6.8

Exercise. The group $A_4$ has no subgroup of order 6.

Proof. Suppose $H$ is a subgroup of $A_{4}$ with order 6. Then it must be isomorphic to $\mathbb{Z}_{6}$ or $S_3$. It is clear that $A_{4}$ has no elements of order 6, since it consists of $3$-cycles and double transpositions, so $H\ncong \mathbb{Z}_6$. Now if $H \cong S_3$, multiplying two distinct elements of order 2 gives an element of order 3. But the only elements of $A_4$ with order 2 are the double transpositions, and multiplying two distinct double transpositions gives another double transposition (which must have order 2). So $H\ncong S_3$, and $A_4$ cannot have a subgroup of order 6.

I.6.10

Exercise. Let $a$ be the generator of order $n$ in $D_n^*$. Show that $\langle a \rangle \triangleleft D_n^*$ and $D_{n}^*/\langle a \rangle \cong \mathbb{Z}_2$.

Proof. It is clear that $\langle a \rangle = \set{e, a, a^{2},…, a^{n-1}}$ has order $n$. Hence, it has index 2 in $D_n^*$, so it must be normal. Moreover, this means that $|D_n^*/\langle a \rangle| = 2$, so we have that $D_n^*/\langle a \rangle \cong \mathbb{Z}_2$.