GRADED HOMEWORK 3
MATH 303, FALL 2013 (WIESNER)
(1) Give an example of a group that has exactly 10 subgroups (including the trivial group and
the group itself.) Hint: Think cyclic groups.
Consider the group Z29 . From the Fundamental Theorem of Cyclic Groups, we know
that the subgroups of Z29 are exactly of the form < k > for each distinct divisor of k.
The distinct divisors of 29 are 20 , 21 , 23 , . . . 29 . Since there are ten such divisors, this is an
example of a group with exactly 10 subgroups.
(2) Let p be an odd prime. Prove there is no group that has exactly p elements of order p.
Proof. Let G be a group. If G has no elements of order p, then we are done. Therefore,
assume that G has at least one element of order p, say g ∈ G. According to Theorems 4.1
and 4.2, < g >= {e, g, g 2 , . . . , g p−1 } and each of the non-identity elements in this cyclic
group, of which there are p − 1, has order p. If G has at least one other element h of order p
not contained in < g >, then the same argument implies that h, . . . , hp−1 are p−1 additional
elements of order p. (Note that there cannot be any non-identity elements that are in both
< g > and < h >. Suppose there were: say g i = hj for some 1 ≤ i, j < p. Since there is
some integer s such that sj ≡ 1modp, we could then argue that h = (hj )s = (g i )s ∈< g >,
a contradiction.) Therefore, G cannot have exactly p elements of order p.
(3) Let α, β ∈ Sn . Prove that αβα−1 and β are both even or both odd permutations.
Proof. Suppose that α = γ1 · · · γr and β = δ1 · · · δs for some transpositions γ1 , . . . , γr , δ1 , . . . , δs .
Then,
αβα−1 = γ1 · · · γr δ1 · · · δs γr · · · γ1 .
Note that s and 2r + s have the same parity, which implies that if β is even then αβα−1 is
even and if β is odd then αβα−1 is odd.
(4) Let α = (1 2 4 6 9)(3 10 8)(5 7). For which m is αm a 5-cycle?
Since disjoint cycles permute (Theorem 5.2), we know that αm = (1 2 4 6 9)m (3 10 8)m (5 7)m .
Note that each individual cycle has prime length and thus prime order. Therefore, (by Theorems 4.2 and 5.3), we know that
• (1 2 4 6 9)m is the identity if 5 | m and a 5-cycle otherwise;
• (3 10 8)m is the identity if 3 | m and a 3-cycle otherwise;
• (5 7)m is the identity if 2 | m and a 2-cycle otherwise.
Therefore, αm is a 5-cycle exactly when 2 | m, 3 | m, and 5 6| m.
(5) Prove that An is a subgroup of Sn for all n.
Proof. We’ll use the two step subgroup test. First, note that the identity e ∈ An (since the
identity is even) and thus An is nonempty.
To show that An is closed, let α, β ∈ An . Then we can write α = γ1 · · · γr and β = δ1 · · · δs
for some transpositions γ1 , . . . , γr , δ1 , . . . , δs , where both r and s are even. From this, we
get that
αβ = γ1 · · · γr δ1 · · · δs .
Date: Due: Oct 11.
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MATH 303, FALL 2013 (WIESNER)
Thus, αβ can be written as a product of r + s transpositions and r + s is even. Therefore,
αβ ∈ An .
To show inverses, again assume α ∈ An and α = γ1 · · · γr . Then, α−1 = α = γr · · · γ1 .
In particular, α−1 can be written as the same number of transpositions as α and so α−1 ∈
An .
(6) How many elements of order 4 does S6 have? How many elements of order 4 does A6 have?
Justify your answer.
From Theorem 5.3, we know that the elements of order 4 is Sn must contain at least one
4-cycle and may contain 2-cycles, but no other cycle types. Thus we have the following
element types:
• (4): There are 64 3! = 90 such cycles. (The first term in the product computes the
number of ways to choose 4 elements from 6. The second term computes the number of
ways that these elements can be arranged in a cycle, always written with the smallest
element first.)
• (4)(2): There are 64 3! = 90 such cycles. (For any 4-cycle, there is only one possible
2-cycle that is disjoint from it.)
Therefore, there are a total of 180 elements of order 4 in S6 . Note that elements of the first
type are odd and elements of the second type are even. (This follows from the fact that
any 4-cycle can be written as the product of three transpositions, which also implies that a
permutation of the form (4)(2) can be written as a product of 4 transpositions.) Therefore,
An has 90 elements of order 4.
√
a 2b
| a, b ∈ Z} (under
(7) Let G = {a + b 5 | a, b ∈ Z} (under addition) and G = {
b a
matrix addition). Show that G is isomorphic to G.
Proof. Define φ : G → G by
√
φ(a + b 5) =
a 2b
b a
.
√
√
First√we show that √
this map is one-to-one. Suppose a + b 5, c + d 5 ∈ G are such that
φ(a + b 5) = φ(c + d 5). This means that
a 2b
c 2d
=
.
b a
d c
In order for two matrices to be equal,
√ entries√in the matrices must be equal, so we get that
a = c and b = d. Therefore, a + b 5 = c + d 5.
√
a 2b
To prove that φ is onto, note that for an arbitrary element
∈ G, a + b 5 ∈ G
b a
√
a 2b
and φ(a + b 5) =
. Therefore, φ is onto.
b a
√
√
Lastly, we check that the function is operation-preserving. Let a + b 5, c + d 5 ∈ G.
Then,
√
√
√
φ((a + b 5) + (c + d 5)) = φ((a + c) + (b + d) 5)
(a + c) 2(b + d)
=
(b + d) (a + c)
a 2b
c 2d
=
+
b a
d c
√
√
= φ(a + b 5) + φ(c + d 5).
GRADED HOMEWORK 3
Therefore, φ is an isomorphism.
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