- #1
sutupidmath
- 1,630
- 4
The question is this: How many isomorphisms f are there from G to G' if G and G' are cyclic groups of order 8?
My thoughts:
Since f is an isomorphism, we know that it prserves the identity, so f:e-->e', e identity in G, e' identity in G'.
Also f preserves the order of each element. That is if o(a)=k=>o(f(a))=k
SO, i thought that f will send the el of the same order in G to the corresponding elements of the same order in G'.
Let G=[a], and G'=. so it means that there are 4 el in G that have order 8 ( the generators of G, a, a^3, a^5,a^7), so there are 4 possibilities for these elements, hense by keeping the other el. fixed we would have 4^4 isomorphisms.
But also we have 2 el of order 4, (a^2, and a^6) so there are two possibilities for these elements to be mapped into G' by f, so if the other el are fixed we would have 2^2 mappings.
Does this mean that the total nr of such isomorphisms is 4^4+2^2? Or am i totally on the wrong way?
Any suggestions would be appreciated.
My thoughts:
Since f is an isomorphism, we know that it prserves the identity, so f:e-->e', e identity in G, e' identity in G'.
Also f preserves the order of each element. That is if o(a)=k=>o(f(a))=k
SO, i thought that f will send the el of the same order in G to the corresponding elements of the same order in G'.
Let G=[a], and G'=. so it means that there are 4 el in G that have order 8 ( the generators of G, a, a^3, a^5,a^7), so there are 4 possibilities for these elements, hense by keeping the other el. fixed we would have 4^4 isomorphisms.
But also we have 2 el of order 4, (a^2, and a^6) so there are two possibilities for these elements to be mapped into G' by f, so if the other el are fixed we would have 2^2 mappings.
Does this mean that the total nr of such isomorphisms is 4^4+2^2? Or am i totally on the wrong way?
Any suggestions would be appreciated.