## Algebra Solutions: Hungerford I.2

1) If $f:G \rightarrow H$ is a group homomorphism, then $f(e_G) = e_H$ and $f(a^{-1}) = (f(a))^{-1} \forall a \in G$
Proof-
First we have that $f(e_G) = f(e_Ge_G) = f(e_G)f(e_G)$ so that $f(e_G) = f(e_G)f(e_G)$. Since we are in a group H we can multiply the inverse to both sides, so whatever element $f(e_G)$ happens to be we get:
$f(e_G)(f(e_G))^{-1} = f(e_G)f(e_G)(f(e_G))^{-1}$
$e_H = f(e_G)$ by cancelling.
Next we show that the inverse of $f(a)$ is $f(a^{-1})$.
$f(a^{-1})f(a) = f(a(a)^{-1}) = f(e_G) = e_H$ Since the inverse of an element f(a) is by definition $(f(a))^{-1}$ and the inverse in any group of an element is unique, we must have that:
$f(a^{-1}) = (f(a))^{-1}$

2) A group G is abelian if and only if the map $\phi : G \rightarrow G$ given by $\phi (x) = x^{-1}$ is an automorphism.
Proof-
$(\Rightarrow )$ First assume that G is abelian and suppose $x,y \in G$. We show the map is an automorphism. First, let $x = y$ then we have
$x^{-1}x=x^{-1}y \Rightarrow e = x^{-1}y \Rightarrow y^{-1} = x^{-1} \Rightarrow \phi (x) = \phi (y)$ so the map is well defined.
Next, $\phi (xy) = (xy)^{-1} = y^{-1}x^{-1} = x^{-1}y^{-1} = \phi (x) \phi (y)$ so the map is a homomorphism.
Assume $\phi (x) = \phi (y)$ Then $x^{-1} = y^{-1} \Rightarrow y = x$ so the map is one to one.
For any $y \in G$ we have that $y^{-1} \in G$ since G is a group. Then $\phi (y^{-1}) = (y^{-1})^{-1} = y$ so that the map is surjective. Thus $\phi$is an automorphism.
$(\Leftarrow )$ Assume that $\phi$ is an automoprhism. Show G is abelian. Let $a,b \in G$ By definition of homomorphism:
$\phi (xy) = \phi (x) \phi (y)$ Then by definition of the map we have:
$(xy)^{-1} = x^{-1}y^{-1} \Rightarrow$
$y^{-1}x^{-1} = x^{-1}y^{-1}$ by property of inverse.
Multiplying on both sides we get
$yx = xy$ so that G is abelian.

5) Let S be a nonempty subset of a group G and define and equivalence relation on G by $a \sim b \text{ iff } ab^{-1} \in S$. Show that $\sim$ is an equivalence relation iff $S \leq G$.
Proof-
$( \Rightarrow )$ Assume that $\sim$ is an equivalence relation. Show that S is a subgroup of G. S inherits associativity from G. Since S is nonempty it must contain some element, call it a. Then since we have an equivalence relation $a \sim a \Rightarrow aa^{-1} \in S \Rightarrow aa^{-1} = e \in S$ So S contains the identity. Since we can write $a = ae = ae^{-1} \in S$ we have that $a \sim e$ and since we have symmetry this gives $e \sim a \Rightarrow ea^{-1} \in S \Rightarrow a^{-1} \in S$. This says that for any element a in S we have the inverse element is also in S. Thus S is a subgroup of G.
$( \Leftarrow )$ This direction is done in the book in section 4 of chapter I.

6) A nonempty finite subset of a group G is a subgroup iff it is closed under the product in G.
Proof-
$( \Rightarrow )$ Trivial
$( \Leftarrow )$ We assume that S is closed under the product in G. That is, assume $ab \in S \forall a,b \in S$. Let $a,b \in S$ then we need only show $b^{-1} \in S$. If $b = e$ then we are done, so suppose $b \neq e$. Note that the set $\{b^m | m \in \mathbb{N} \} \subset S$ is finite. Being finite tells us that there exists some numbers $i,j$ with the property $i > j$ such that $b^i = b^j$. Then $b^{i-j} = e$. Then since $b \neq e$ we have that
$i-j \geq 2 \Rightarrow i-j-1 \geq 1$. Then $b^{i-j-1} \in \{b^m | m \in \mathbb{N} \} \subset S$.
Then finally:
$b^{i-j} = e$
$bb^{i-j-1} = e$
$b^{i-j-1} = b^{-1}$ so we have found the inverse of b to be $b^{i-j-1}$ which was shown to be in S. Since the product is closed in S, we have $ab^{-1} \in S$ so by the subgroup test, S is a subgroup of G.

13) If $G = \langle a \rangle$ is a cyclic group and H is any group, then the homomorphism $f:G \rightarrow H$ is completely determined by the element $f(a) \in H$
Proof-
Suppose we know the value of $f(a) = b$. Then for any $x \in G$, x can be written as $x = a^n$ for some integer n, since G is cyclic. Then $f(x) = f(a^n)$ because f is a homomorphism we have:
$f(a^n) = (f(a))^n = b^n$, so what f does to any element x is completely determined by what it does to a.

15) b) $Aut( \mathbb{Z}_p) \cong \mathbb{Z}_{p-1}$ (p prime)
Proof-
Because $\mathbb{Z}_n$ is cyclic, we know from 13) that any homomorphism on $\mathbb{Z}_n$ is determined by where the generating element 1 is sent. So for any map $\phi \in Aut( \mathbb{Z}_n)$ write $\phi _k \text{ iff } \phi (1) = k$. Then we have:
$Aut( \mathbb{Z}_p) = \{ \phi _1, \phi _2,..., \phi _{p-1}\}$. Then the map $f: Aut( \mathbb{Z}_p) \rightarrow \mathbb{Z}_{p-1}$ given by $f(\phi _k) = k$ is an isomorphism.