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.

Advertisements
This entry was posted in Algebra Solutions. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s