Solutions to Assignment 8
Turn this in at the start of recitation on Friday, November 7.
1. Show that Z2 × Z3 × Z4 is not isomorphic to Z24 even though 2, 3, and 4 share no common factor.
There are several ways to do this. For example, if (a, b, c) ∈ Z2 × Z3 × Z4 , then (a, b, c)12 =
(a12 , b12 , c12 ) = (e, e, e), so the order of any element of Z2 × Z3 × Z4 is at most 12, while 1 ∈ Z24
has order 24.
Another way is to note that Z24 has exactly one subgroup of order 2, namely the subgroup consisting
of 0 and 12, whereas Z2 × Z3 × Z4 has two: {(0, 0, 0), (1, 0, 0)} and {(0, 0, 0), (0, 0, 2)}.
2. List all of the finite abelian groups of order 36. Show that all of them have a subgroup of order 6.
By the Fundamental Theorem of Finite Abelian Groups, any abelian group of order 36 is a direct
product of cyclic groups whose order is a power of a prime. Since 36 = 22 32 , these cyclic groups have
orders 2, 4, 3, or 9. The possibilities are Z4 × Z9 , Z2 × Z2 × Z9 , Z4 × Z3 × Z3 , and Z2 × Z2 × Z3 × Z3 .
These all have subgroups of order 6. Namely,
{0, 2} × {0, 3, 6} ⊂ Z4 × Z9
Z2 × {0} × {0, 3, 6} ⊂ Z2 × Z2 × Z9
{0, 2} × {0} × {0, 3, 6} ⊂ Z4 × Z3 × Z3
Z2 × {0} × Z3 × {0} ⊂ Z2 × Z2 × Z3 × Z3 .
3. Suppose that G is an abelian group of order n and that m is a factor of n. Show that G has a subgroup
of order m.
As in the hint, the Fundamental Theorem of Finite Abelian Groups implies that we can write G =
Zn1 × · · · × Znk , where n = n1 . . . nk . Since m divides n, we can write m = m1 . . . mk where mi divides
ni for all i.
For each i, let Hi = hni /mi i ∈ Zni so that Hi is a subgroup of order mi . Then
H1 × · · · × Hk = {(a1 , . . . , ak ) | ai ∈ Hi }
is a subgroup of G of order m1 . . . mk = m, as desired.
4. Give an example of a group G and normal subgroups H1 , H2 , and H3 such that Hi ∩ Hj = {e} when
i 6= j, but G is not isomorphic to H1 × H2 × H3 .
(There was a missing condition in this problem – there should have been the condition that G = H1 H2 H3
as well.)
Let G = R2 . Let
H1 = {(t, 0) | t ∈ R}
H2 = {(t, t) | t ∈ R}
H3 = {(0, t) | t ∈ R}.
These are all subgroups, and H1 + H2 + H3 = G. (Indeed, H1 + H3 = R2 because (x, y) = (x, 0) + (0, y)
for all x, y ∈ R.) On the other hand,
H1 ∩ H2 = H2 ∩ H3 = H1 ∩ H3 = {(0, 0)}.
5. Verify Sylow’s theorem for S4 by finding subgroups of orders 2, 4, and 8.
Order 2 H = {e, (12)}
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Order 4 The permutation (1234) has order 4, so its powers form a cyclic subgroup of order 4.
Order 8 Let a = (1234) and b = (14)(23). We claim that
H = {e, a, a2 , a3 , b, ba, ba2 , ba3 }
= {e, (1234), (13)(24), (4321), (14)(23), (24), (12)(34), (13)}
is a subgroup of S4 . Here are some ways to show this – can you find some others?
• Calculate a multiplication table. This is the way that takes the most calculation.
• Find a formula for multiplying a and b. For instance, ba = a−1 b, a4 = e, and b2 = e, so if
j, k ∈ Z, then
baj bak = b2 a−j ak = ak−j .
Similarly,
aj bak = b2 a−j ak = bak−j
aj ak = aj+k
baj ak = baj+k .
The right sides of all these formulas are elements of H, so H is closed under multiplication.
• Use symmetry. Label the corners of a square 1, 2, 3, and 4:
4
2
1
3
Any symmetry of the square corresponds to a permutation of the vertices. Counterclockwise
rotation by 90◦ corresponds to the permutation (1234) and flipping across the horizontal
axis corresponds to the permutation (14)(23). There are eight symmetries of the square,
corresponding to the eight permutations in H.
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