Thread: Permutation Groups

Hi everyone,

I could really use some help with this problem:

Let H be the subset of S₃ consisting of all the permutations that leave the element 2 fixed.

(i) Prove that H is a subgroup.
(ii) Prove that H is isomorphic to S₂

NOTE: I am unfamiliar with this site, so my work is not written in standard notation. I am sorry it's difficult to read, but thank you for bearing with me.

I began the problem by writing out all of the permutations contained in S₃ and found that the two permutations which kept 2 fixed were the identity permutation, p₀ = (1 2 3 /1 2 3) and another permutation m₂ = (1 2 3 /3 2 1). Therefore H = {p₀, m₂}

(i) To start the proof, I wrote out the group table for H and noted that it was closed.
Then for the identity:
p_{0 *} p_{0 =} p₀
m_{2 *} p_{0 =} m_{2
Therefore, the identity element,} p<sub>0, is also in H

Then for the inverse:</sub>
p_{0 *} p_{0 =} p₀
m_{2 *} m₂ = p₀
Therefore, every element has an inverse

Thus, H is a subgroup of S₃

(ii) This part of the problem was much trickier for me...
I defined S₃ as {e = (1 2 /1 2), a = (1 2 /2 1)}

I then defined a function from H to S2 such that identities were mapped to identities (f(Po) ---> e) and the remaining two permutations were mapped to each other (f(m2) ---> a). As a result, I had a bijective function. Therefore, the only thing left to find is a homomorphism.

Using the above defined function, I computed:
f(p0 * m2) = f(p0) * f(m2)
f(m2) = e * a
a = a

f(m2 * p0) = f(m2) * f(p0)
f(m2) = a * e
a = a

Thus, a homomorphism.
Therefore, H is isomorphic to S2.

Is this a valid proof?

Thank you for any guidance you can provide me!

To prove that it is a homomorphism, you have to check for the product m2*m2 too. For the product e*e is trivial, but in a test I would put it. The rest is correct.

you can actually prove a STRONGER result: any two groups of order 2 are isomorphic. let G = {e,a} and G' = {e',b} be two groups of order 2.

we will show that φ:G→G' given by: φ(a^k) = b^k for k = 0,1 is an isomorphism.

first, note that by definition, φ(e) = φ(a⁰) = b⁰ = e'.

also, note that since G is a group, a² must be an element of G, and since a² = a implies a = e (and a and e are DIFFERENT),

it must be the case that a² = e. similarly reasoning show that b² = e'.

so in G we have:

e*e = e
a*e = a
e*a = a
a*a = e

and in G' we have:

e'*e' = e'
b*e' = b
e'*b = b
b*b = e'

it is clear that φ is bijective, and we could argue that since G and G' have "virtually" the same multiplication table, they are isomorphic.

but let's use the definition of a homomorphism. to do so, we need a more compact description of the multiplication in G and G'. and that is fairly easy:

(a^k)(a^m) = a^{(k+m mod 2)}, and (b^k)(b^m) = b^{(k+m mod 2)} (you can verify this holds, there are only 4 products to check).

so φ(a^ka^m) = φ(a^{(k+m mod 2)}) = b^{(k+m mod 2)} = (b^k)(b^m) = φ(a^k)φ(a^m).

thus φ is a homomorphism.

now, we simply observe that both {e, (1 3)} and {e, (1 2)} are groups of order 2.

(here (1 3) is the map:

1→3
2→2
3→1 of S₃, and

(1 2) is the map:

1→2
2→1 of S₂, just to clarify).

in fact, you can even generalize THIS result (although i won't prove it here). if p is ANY positive prime integer, then any two groups of order p are isomorphic. so in light of this, it merely suffices to note that 2 is a prime number.