Abstract Algebra I
Practice Problems (Wed)
1. Suppose a group G of order 35 acts on a set S of size 9. Are there any fixed points? Prove
your answer. [Hint: First, write down what sizes the orbits can be]
Solution: Yes, there are at least two fixed points. Recall that an object x ∈ S is a fixed
point if and only if the orbit of x consists of only x.
Proof: By the Orbit-Stabilizer Theorem, the size of the orbit must divide 35, so the size
of an orbit can be 1, 5, or 7. So the possible orbit sizes are the following:
9 orbits, each of size 1.
one orbit of size 5, and four orbits (each of size 1).
one orbit of size 7, and two orbits (each of size 1).
2. Let H be a subgroup of G and let x ∈ G. You can use, without proof, a previous result (from
egunawan.github.io/algebra/slides/sec3p3.pdf) that the set xHx
−1 is a subgroup of G.
(a) Prove that H ∼ = xHx−1 . [Hint: Define a map and show it is (i) a homomorphism, (ii)
injective, and (iii) surjective.]
Solution: Define f : H → xHx−1 by f (h) = xhx−1 .
(i) To show that f is a homomorphism, let a, b ∈ H. Then
f (ab) = xabx−1
= xax−1 xbx−1
= f (a)f (b).
(ii) To show that f is injective, let f (a) = f (b). Then xax−1 = xbx−1 .
First multiply both sides on the right by x, then multiply both sides on the left by
x−1 .
(iii) To show that f is onto, let y ∈ xHx−1 . This means that y = xhx−1 for some
h ∈ H.
Then f (h) = xhx−1 = y.
(b) Prove that if H is the unique subgroup of G of (finite) order |H|, then H must be normal.
You may use the result of Part (a) even if you didn’t prove it.
Solution: By Part (a), xHx−1 ∼
= H for all x ∈ G, so |H| = |xHx−1 | for all x ∈ G.
Since H is the unique subgroup of order |H|, we must have H = xHx−1 for all
x ∈ G. So H C G.
Practice Problems (Wed)
1. Suppose a group G of order 35 acts on a set S of size 9. Are there any fixed points? Prove
your answer. [Hint: First, write down what sizes the orbits can be]
Solution: Yes, there are at least two fixed points. Recall that an object x ∈ S is a fixed
point if and only if the orbit of x consists of only x.
Proof: By the Orbit-Stabilizer Theorem, the size of the orbit must divide 35, so the size
of an orbit can be 1, 5, or 7. So the possible orbit sizes are the following:
9 orbits, each of size 1.
one orbit of size 5, and four orbits (each of size 1).
one orbit of size 7, and two orbits (each of size 1).
2. Let H be a subgroup of G and let x ∈ G. You can use, without proof, a previous result (from
egunawan.github.io/algebra/slides/sec3p3.pdf) that the set xHx
−1 is a subgroup of G.
(a) Prove that H ∼ = xHx−1 . [Hint: Define a map and show it is (i) a homomorphism, (ii)
injective, and (iii) surjective.]
Solution: Define f : H → xHx−1 by f (h) = xhx−1 .
(i) To show that f is a homomorphism, let a, b ∈ H. Then
f (ab) = xabx−1
= xax−1 xbx−1
= f (a)f (b).
(ii) To show that f is injective, let f (a) = f (b). Then xax−1 = xbx−1 .
First multiply both sides on the right by x, then multiply both sides on the left by
x−1 .
(iii) To show that f is onto, let y ∈ xHx−1 . This means that y = xhx−1 for some
h ∈ H.
Then f (h) = xhx−1 = y.
(b) Prove that if H is the unique subgroup of G of (finite) order |H|, then H must be normal.
You may use the result of Part (a) even if you didn’t prove it.
Solution: By Part (a), xHx−1 ∼
= H for all x ∈ G, so |H| = |xHx−1 | for all x ∈ G.
Since H is the unique subgroup of order |H|, we must have H = xHx−1 for all
x ∈ G. So H C G.
