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About LinkBacksTo 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.
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:
p0 * p0 = p0
m2 * p0 = m2
Therefore, the identity element, p0, is also in H
Then for the inverse:
p0 * p0 = p0
m2 * 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!
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(ak) = bk 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:
(ak)(am) = a(k+m mod 2), and (bk)(bm) = b(k+m mod 2) (you can verify this holds, there are only 4 products to check).
so φ(akam) = φ(a(k+m mod 2)) = b(k+m mod 2) = (bk)(bm) = φ(ak)φ(am).
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.
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