# Algebra Questions

1. Dec 23, 2006

### acm

Two unrelated question I need checked.
i) Every C-Cycle, C>1, can be written as a product of transpositions. Every permutation of a finite number n>1 of elements can be expressed as a product of transpositions.
ii) (Cayley's representation theorem for groups) For every group G is isomorphic to a group of permutations A [G]=n then G is isomorphic to a group of permutations on objects.

i) a) (Proof of C >1) Every C cycle can be represented in the form (c_2,c_1), (c_3,c_1),...(c_i,c_1),....,(c_c,c_1). Hence this defines the mapping from c_1 to c_c, thus in general it suffices to say that 1 <= i < C. As there is a mapping from c_1 to c_c such that every other number in the group remains unchanged, then i=1, thus that 1=n. Hence 1 < C.
b) (Proof of n > 1) Hence from a), any C-Cycle with C>1 can be written as a product of transpositions. Thus as permutations can be represented as C cycles on a arbitary set then n > 1 is satisfied by part a) #.

ii) (Proof of one to one correpondence) Let the set A be defined as (p_1, p_2, .... , p_n,...) Hence, let A undergo the mapping p_1 -> g_1 such that the image of A under the mapping is: (p_1g_1, ..., p_ng_n,....). Thus there is an equivalence between p_n ~ (p_1g_1,....., p_ng_n).
(Proof of Isomorphism) For an isomorphism to be acting from A to the permutation set then associotivity must hold. Let p_k = (p_1g_1,....., p_1g_k), Hence if G is isomorphic to the permutation of G then p_k(p_n * c) = (p_k * p_n)c. p_k * p_n = ([p_1,.....,p_k],[p_1g_1,.....,p_kg_k,...]) * (([p_1,.....,p_n],[p_1g_1,.....,p_ng_n,....]). As the permutation group has the same binary operation as the group itself. Then there is an isomorphism between the group G and the group of permutations on the group.

I'm having trouble understanding the concept of an isomorphism, the book I'm working from defines it as a special kind of equavlence which acts on the relation between two groups, whether that is binary, etc. So the group may have different contents but if the operation holds then there is an isomorphism. Is this correct or am I doing something wrong?

2. Dec 25, 2006

### Swapnil

This question is way too complicated to be in pre-calculus forum. "Set Theory, Logic, Probability, Statistics" forum is a far better forum for these types of questions. I would ask the moderators to please move this question over there.

3. Dec 25, 2006

### HallsofIvy

Staff Emeritus
Since this is, still, a homework question, I am moving it to "Calculus and beyond".

4. Dec 25, 2006

### HallsofIvy

Staff Emeritus
I don't understand what you mean by "acts on the relation between two groups". What relation? What is it that is "binary, etc."? (The group operation is always binary.) An isomorphism between two groups is a function from one group to another that "preserves" the operation (f(a*b)= f(a)*f(b)) and invertible. Basically, it means the two groups are really just different "labelings" of the same thing. That is if f(a)= z, f(b)= y, f(c)= z, etc. and you just "relabel" "a" to be "z", "b" to be "y" etc. then the two groups will look exactly the same. One of the things that means is that while a group may have "different contents" (by which I presume that you mean the underlying sets are different) they must have exactly the same number of elements.
What "operation" are you talking about? The group operation? And what do you mean by "holds"? If you mean that f(a*b)= f(a)o f(b) where "*" is the operation in one group and "o" is the operation in the other then that defines a homomorphism, not an isomorphism since it is not necessarily invertible (one-to-one and onto).
For example, let G be the group with elements {x, y} and operation defined by x*x= x, x*y= y, y*a= y, y*y= x (so that x is the identity and y is its own inverse. Let H be the "rotation" group of order 4 with elements {a, b, c, d} and operation defined by aoa= a, aob= b, aoc= c, aod= d, boa= b, bob= c, boc= d, bod= a, coa= c, cob= d, coc= a, cod= b, doa= d, dob= a, doc= b, dod= c. The a is the identity, b and d are inverses, and c is its own inverse.

The function f(x)= a, f(y)= c is a homomorphism but not an isomorphism because it is not "onto": nothing is mapped into either b or d. Conversely, the function g(a)= x, g(b)= x, g(c)= y, g(d)= x is also a homomorphis but not an isomorphism because it is not "one-to-one". More than one element is mapped to x. If we restrict g to the subgroup {a, c} then g(a)= x, g(c)= y is an isomorphism and its inverse is f.