Homework 3 comments
1. Artin 2.5.1. There was a little confusion about the definition of a cyclic
group. A group G is cyclic if it is generated by a single element, i.e. there
is some x ∈ G (a “generator”) such that G = {xm : m ∈ Z}. The correct
solution to G cyclic implies G0 cyclic is: Let x be a generator of G. Then
ϕ(x) is a generator of G0 . To see this, let y 0 be any element of G0 . By
surjectivity, there exists y ∈ G such that ϕ(y) = y 0 . Any such y equals xm
for some m ∈ Z (since G is cyclic), so y 0 = ϕ(x)m , i.e. ϕ(x) generates G0 .
2. Transpositions generate Sn . I’ll explain three proofs:
• Any element of Sn can be expressed as a product of cycles, so it suffices to check that any cycle (i1 i2 . . . ik ) is a product of transpositions.
But we can see this by direct calculation:
(i1 i2 . . . ik ) = (i1 i2 )(i2 i3 ) · · · (ik−1 ik ).
(Alternatively,
(i1 i2 . . . ik ) = (i1 ik )(i1 ik−1 ) · · · (i1 i2 ),
so this decomposition into a product of transpositions is certainly not
unique.)
• By induction. Clear for n = 2. Assume the result is established for
Sn−1 , and consider any element σ ∈ Sn . There are two cases: σ
either does or does not fix some element of {1, 2, . . . , n}. In the first
case, supposed σ(i0 ) = i0 . The subgroup {f ∈ Sn : f (i0 ) = i0 } is
isomorphic to Sn−1 , via the map that sends f to its restriction to
{1, . . . , n} \ {i0 } (a set with n − 1 elements). Applying the induction
hypothesis, we win. Now we consider the case where σ ∈ Sn fixes no
element of {1, . . . , n}. Consider the permutation τ = σ · (σ −1 (n), n)
(that is, precompose σ with this transposition). By construction,
τ (n) = n, so from the previous argument we conclude that τ is a
product of transpositions; but then visibly σ is as well.
• (Solution using row reduction). Recall the injective homomorphism
p : Sn → GLn (C), whose image is precisely the subgroup of permutation matrices. By construction, left-multiplication by p ((ij)) is the
row-operation that swaps the i and j rows. Now consider any permutation matrix P = p(σ). We can row-reduce P to the identity using
only row-operations of the form “swap two rows,” hence we can write
p(σ) = P = p ((i1 j1 )) · · · p ((ik jk ))
for some indices i1 , . . . , ik and j1 , . . . , jk . Since p is injective, we
deduce that
σ = (i1 j1 ) · · · (ik jk ).
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Finally, note that you don’t need all the transpositions to generate Sn :
you only need those of the form si = (i, i + 1), 1 ≤ i ≤ n − 1. This follows
from the relation
(ij) = si si+1 · · · sj−2 sj−1 sj−2 · · · si+1 si .
This innocent-looking calculation is actually quite important!
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