Solutions to Practice Exam 4
Modern Algebra
Wednesday, November 23, 2011
Show all your work and justify all your answers. Be sure to write your proofs in complete
sentences, whether in English words or symbolic notation. If you define a function or operation based on cosets, be sure to show that it is well-defined. If you define a homomorphism,
be sure to show that it is homomorphic. Make your examples concrete and specific.
1. Give an example of each of the following:
[15 points]
(a) A domain that is not a field. Solution. Z.
(b) A finite field. Solution. Z7 .
(c) A commutative ring without unity. Solution. 2Z.
(d) A ring with zero divisors. Name two of its zero divisors.
Solution. Z6 has zero divisors [2] and [3].
2. Definition 1 Let R be a ring and let I ⊆ R. Then I is called an ideal of R if
(a) I is closed under addition and taking negatives, and
(b) for every u ∈ I and every r ∈ R, ur ∈ I and ru ∈ I.
Let F be a field. Prove that the only ideals of F are {0} and F .
[10 points]
Proof. Suppose I is an ideal of F that is neither {0} nor F . Then I contains some
element 0 6= a ∈ F . Since a ∈ I and a−1 ∈ F , we conclude that 1 = aa−1 ∈ I. Now let
x ∈ F ; then since 1 ∈ I, we have x = x1 ∈ F . Thus F ⊆ I so I = F , a contradiction.
We conclude that F has no ideals except {0} and F .
3. Let R be a commutative ring with unity, and let a ∈ R be a zero divisor. Prove that
a is not invertible.
Proof. Suppose a is invertible; then there exists b ∈ R such that ab = e. Since a is
a zero divisor, there exists a nonzero c ∈ R such that ca = 0. Then multiplying the
equation ab = e on the left by c, we have cab = ce, which implies 0b = c so 0 = c, a
contradiction. Thus a is not invertible.
4. Give a specific example of a group G and a subgroup H of G such that H is not normal
in G. Be sure to demonstrate that H 6 / G.
[10 points]
Solution. Of course there are infinitely many correct answers.
For example, {R0 , V } 6 / D8 and {(1), (12)} 6 / S3 .
5. Let G be a group, and let both N and K be normal subgroups of G. Let
N K = {nk : n ∈ N, k ∈ K}.
Practice Exam 4 Solutions
Page 2 of 3
Modern Algebra, Fall 2010
It is a fact that N K is a subgroup of G. Prove that N K is normal in G. [10 points]
Proof. Let x ∈ N K and let g ∈ G. Then x = nk for some n ∈ N and some k ∈ K.
Then
gxg −1 = g(nk)g −1 = gnekg −1 = gn(g −1 g)kg −1 = (gng −1 )(gkg −1 ) ∈ N K
since N / G and K / G. Thus N K / G as desired.
6. In S4 , the subgroup N = {(1), (12)(34), (13)(24), (14)(23)} is normal.
[15 points]
(a) List the elements of the group S4 /N .
Solution. The elements of S4 /N are the six cosets of N in S4 , namely
N
= { (1) , (12)(34), (13)(24), (14)(23)}
N (12) = { (12) , (34) , (1423) , (1324) }
N (13) = { (13) , (1432) , (24) , (1234) }
N (14) = { (14) , (1342) , (1243) , (23) }
N (123) = {(123), (243) , (142) , (134) }
N (132) = {(132), (234) , (124) , (143) }.
(b) Is S4 /N abelian? Justify your answer. Solution. No, it is not. We need only
produce two noncommuting elements of S4 /N , such as N (12) and N (123):
N (12) N (123) = N ((12)(123)) = N (23) = N (14)
N (123) N (12) = N ((123)(12)) = N (13).
Since N (12) N (123) 6= N (123) N (12), we conclude S4 /N is nonabelian.
7. Let G be a group, let N and H be subgroups of G, and suppose N / G. Let
N H = {nh : n ∈ N, h ∈ H}.
It is a fact that N H is a subgroup of G; moreover, N / N H.
You also proved in a homework assignment that N ∩ H / H.
Prove that H/(H ∩ N ) ∼
= N H/N .
(Suggestion: Define a function θ(h) = N h and use the Fundamental Homomorphism
Theorem.)
[10 points]
Proof. Let θ : H → N H/N be defined by θ(h) = N h. Then θ is clearly onto, and θ is
a homomorphism since θ(h1 )θ(h2 ) = (N h1 )(N h2 ) = N (h1 h2 ) = θ(h1 h2 ). We see that
h ∈ ker θ ⇐⇒ θ(h) = N ⇐⇒ N h = N ⇐⇒ h ∈ N,
so ker θ = {h ∈ H : h ∈ N } = H ∩ N . Finally, the Fundamental Homomorphism
Theorem tells us
N H/N ∼
= H/ ker θ = H/(H ∩ N )
as desired.
Practice Exam 4 Solutions
Page 3 of 3
Modern Algebra, Fall 2010
8. Let G = Z × Z, and define θ : G → Sym(8) by
[15 points]
θ((a, b)) = (123)a (45)b .
(a) Find ker θ. Solution. ker θ = {(3m, 2n) : m, n ∈ Z}.
(b) Find θ(G) and list all its elements.
Solution. θ(G) = {(1), (123), (132), (45), (123)(45), (132)(45)}.
(c) Compute [G : ker θ]. Justify your answer.
Solution. By the Fundamental Homomorphism Theorem,
[G : ker θ] = |G/ ker θ| = |θ(G)| = 6.
XC. Let G be a group of order 105 and H a group of order 98. Suppose θ : G → H is a
homomorphism, and assume there exists some a ∈ G such that θ(a) 6= eH . Find |θ(G)|
and | ker θ|.
[10 bonus points]
Solution. Since θ(G) is a subgroup of H, we know |θ(G)| 98. Since θ(G) ∼
= G/ ker θ,
we know |θ(G)| = |G|/| ker θ| so |G| = |θ(G)| · | ker θ|. Thus |θ(G)| 105.
Putting these two pieces together yields |θ(G)| gcd(98, 105) = 7, so |θ(G)| is either
1 or 7. Since θ(a) 6= eH for some a ∈ G, we know |θ(G)| =
6 1, so |θ(G)| = 7. Thus
| ker θ| = 105/7 = 15.