## Algebra Solutions: Hungerford I.1

1) Give examples of semigroups and monoids which are not groups
Solution- $(\mathbb{N},+)$ is a semigroup, but neither a monoid nor a group. This is because the number 0 is not in the natural numbers, thus there is no additive identity.
$(\mathbb{N},\cdot)$ is a monoid because 1 is the identity, however it has no inverses.

2) Let G be a group, S a nonempty 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-
We must check that the operation is associative. We have that $(f+(g+h))(s) = f(s) + (g+h)(s) = f(s) + g(s) + h(s) = (f + g)(s) + h(s) = ((f+g)+h)(s)$
Now, G is a group so it has an identity element by definition, call it 0. Define a function $e:S \rightarrow G$ by $e(s) = 0$ $\forall s \in S$. This is by definition a function so it belongs to M (S,G). Then we have $\forall f \in M (S,G), (e+f)(s) = e(s) + f(s) = 0 + f(s) = f(s)$ so that e is the identity of M (S,G).
Lastly, define for any function $f \in M (S,G)$ the function $g:S \rightarrow G$ defined by $g(s)= -f(s) \forall s \in S$. Then we have $(g+f)(s) = g(s) + f(s) = -f(s) + f(s) = 0 = e(s)$.
If G is abelian then we have $(f+g)(s) = f(s) + g(s) = g(s) + f(s) = (g+f)(s) \forall f,g \in M(S,G)$ Therefore M(S,G) is abelian.

7) If p is prime, then the nonzero elements of $\mathbb{Z}_p$ form a group of order p-1 under multiplication (mod p).
Proof-
$\mathbb{Z}_p \backslash \{0\}$ inherits associativity from multiplication of integers, as well as the identity element 1. Since p is prime, for any $a \in \mathbb{Z}_p \backslash \{0\}$ we have that gcd(a,p) = 1. Thus, there exists $k_1, k_2 \in \mathbb{Z}$ such that $k_1a+k_2p = 1$. Then by doing mod p on both sides we have $k_1a = 1$ (mod p), so that $a^{-1} = k_1$. Thus we have an inverse for each element and hence a group of order p-1 elements.

8) a) The relation given by $a \sim b \leftrightarrow a - b \in \mathbb{Z}$ is a congruence relation on the additive group $\mathbb{Q}$
Proof-
Let $a,b,c,d \in \mathbb{Q}$. First, since $a-a=0 \in \mathbb{Z}$ we have that $\sim$ is reflexive. If $a \sim b$ then $a-b = k$ for some $k \in \mathbb{Z}$ so then $b - a = -k \in \mathbb{Z}$ so that the relation is symmetric. If $a \sim b$ and $b \sim c$ then we have $a-b = m$ and $b - c = n$ for some integers, m and n. Then putting them together we have $a-c = b+m+n-b = m+n \in \mathbb{Z}$ so the relation is transitive. Lastly, suppose that $a \sim b$ and $c \sim d$. Then $a-b = r$ and $c-d=s$ for some integers r and s. To show congruence we must show that $(a+c) \sim (b+d)$. Consider $(a+c) - (b+d) = ((b+r)+(d+s))-((a-r)+(c-s))= 2r+2s+b-a+d-c=2r+2s-r-s = r+s \in \mathbb{Z}$ Thus it is a congruence relation.

b) The set $\mathbb{Q} \backslash \mathbb{Z}$ of equivalence classes is an infinite abelian group.
Proof-
From part a) we know that the relation is a congruence relation. Theorem 1.5 in Hungerford gives us that the set of equivalence classes of an abelian group under a congruence relation is an abelian group.

9) Let p be a fixed prime. Let $R_p$ be the set of all those rational numbers whose denominator is relatively prime to p. Let $R^p$ b the set of rationals whose denominator is a power of p. Prove both $R_p$ and $R^p$ are groups under ordinary addition.
Proof-
First we consider $R_p$ and we show closure. Let $\frac{a}{b}, \frac{c}{d} \in R_p$. Then $\frac{a}{b} + \frac{c}{d} = \frac{ad+bc}{bd}$ We have that $gcd(b,p)=1=gcd(d,p)$ so that $p \nmid b$ and $p \nmid d$ which gives that $p \nmid bd$. Thus $gcd(bd,p) = 1$ so the operation is closed.
Associativity is inherited from the ordinary addition of rationals.
The identity 0, can be written as $\frac{0}{p}$ where p is some other prime so that $gcd(p,p`)= 1$.
The inverse of an element $\frac{a}{b}$ is $\frac{-a}{b}$
$R^p$ is very similar except easier.

11) The following conditions on a group G are equivalent:
(i) G is abelian
(ii)$(ab)^2 = a^2b^2 \forall a,b \in G$
(iii)$(ab)^{-1} = a^{-1}b^{-1} \forall a,b \in G$
(iv)$(ab)^n = a^nb^n \forall n \in \mathbb{Z} , \forall a,b \in G$
(v) $(ab)^n = a^nb^n$ for 3 consecutive integers n and $\forall a,b \in G$
Proof-
$(i) \Rightarrow (iv)$ Let G be abelian. The case when n is 0 is trivial. Let $n \geq 1$ Then $(ab)^1 =a^1b^1$ Assume it is true for some natural number k and then use induction. If $n \leq 1$ then we have $-n \geq 1$ so we can use the previous case to show it is true.
$(iv) \Rightarrow (v)$ follows immediately.
$(v) \Rightarrow (i)$ Let $a,b \in G$. Assuming for some n the hypothesis is true, we have $(ab)^n = a^nb^n$, $(ab)^{n+1} = a^{n+1}b^{n+1}$, $(ab)^{n+2} = a^{n+2}b^{n+2}$. We use these facts to say first that
$(ab)^{n+1} = a^{n+1}b^{n+1}$
$(ab)^n(ab) = a^{n}ab^{n}b$
$a^nb^nab = a^{n}ab^{n}b$
$b^na = ab^{n}$
We also have
$(ab)^{n+2} = a^{n+2}b^{n+2}$
$(ab)^n(ab)^2 = a^{n}a^2b^{n}b^2$
$a^nb^nabab = a^{n}aab^{n}bb$
$b^naba = aab^{n}b$
Then using the fact that $b^na = ab^{n}$, we get
$ab^nba = ab^{n}ab$
$ba = ab$ So that G is abelian.
$(i) \Rightarrow (ii)$ Trivial
$(ii) \Rightarrow (iii)$ Assume that $(ab)^2 = a^2b^2 \forall a,b \in G$. Then we have:
$(a^{-1}b^{-1})^2 = (a^{-1})^2(b^{-1})^2$
$a^{-1}b^{-1}a^{-1}b^{-1} = a^{-1}a^{-1}b^{-1}b^{-1}$
$b^{-1}a^{-1} = a^{-1}b^{-1}$
$(ab)^{-1} = a^{-1}b^{-1}$
$(iii) \Rightarrow (i)$ Assume that $(ab)^{-1} = a^{-1}b^{-1}$. Then
$(ab)^{-1} = a^{-1}b^{-1}$
$(ab)^{-1} = (ba)^{-1}$
$e = (ba)^{-1}ab$
$ba = ab$ so that G is abelian.

13) If $a^2 = e \forall a \in G$, then G is abelian.
Proof-
The fact that $a^2 =e$ gives us that $a = a^{-1} \forall a \in G$. Let $a,b \in G$. Then we know $(ab)^2 = e = abab = a^{-1}b^{-1}ab$
$\Rightarrow e =a^{-1}b^{-1}ab$
$\Rightarrow ba =ab$
This is true for any arbitrary elements a and b, so G is abelian.

14) If G is a finite group of even order then G contains an element $a \neq e$ such that $a^2 = e$.
Proof-
Suppose that no element satisfies the given condition. So that there is no element a such that $a = a^{-1}$. Then each element can be paired with it’s inverse to give 2n elements in G for some n. All of these pairs together with the identity element e, gives 2n + 1 elements in G, which is a contradiction to G being of even order. Therefore such an element exists.