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.
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).
\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}
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.
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.
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
(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.
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.
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.

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