
Permutation Groups
Hi everyone,
I could really use some help with this problem:
Let H be the subset of S_{3} 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 S_{2 }
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 S_{3} and found that the two permutations which kept 2 fixed were the identity permutation, p_{0} = (1 2 3 /1 2 3) and another permutation m_{2} = (1 2 3 /3 2 1). Therefore H = {p_{0}, m_{2}}
(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 = }p_{0 }m_{2 * }p_{0 }_{= }m_{2 }_{Therefore, the identity element, }p_{0, is also in H }_{
Then for the inverse: }p_{0 * }p_{0 = }p_{0 }m_{2 * }m_{2 }= p_{0 }Therefore, every element has an inverse
Thus, H is a subgroup of S_{3}
(ii) This part of the problem was much trickier for me...
I defined S_{3} 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!

Re: Permutation Groups
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.

Re: Permutation Groups
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(a^{0}) = b^{0} = e'.
also, note that since G is a group, a^{2} must be an element of G, and since a^{2} = a implies a = e (and a and e are DIFFERENT),
it must be the case that a^{2} = e. similarly reasoning show that b^{2} = 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^{k}a^{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 S_{3}, and
(1 2) is the map:
1→2
2→1 of S_{2}, 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.