Elements whch correspond under an isomorphism.

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

Yes, i understand the problem, but i was just asking out of curiosity if the neutral element had any significance in determining an isomorphism.

Shadow · **Posted:** Mon, 5 Mar 2012 21:51:51 UTC

DgrayMan wrote:

Yes, i understand the problem, but i was just asking out of curiosity if the neutral element had any significance in determining an isomorphism.

Ah, then I'm sorry. It seemed as if you had a fixation on trying to do it that way.

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

I know I'm not the brightest crayon in the box. I understand :/

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

For the question, if G1 is a cyclic group w/ generator a , prove that G2 is also a cyclic group with generator f(a).
i said:
Since f:G1->G2, is an isomorphism and G1 and G2 are cyclic, then G1 and G2 have the same order n , (some n in Z). Therefore since they have the same order, f(a)=a is also cyclic.//

Teacher said I'm trying to show G2 has order n, and that i should use if G1=<a> i have to prove G2=<f(a)>

So since G1=<a> and f:G1->G2 is isomorphic, then a=f(a)
and G2=<a>=<f(a)>

Shadow · **Posted:** Mon, 26 Mar 2012 04:41:21 UTC

DgrayMan wrote:

For the question, if G1 is a cyclic group w/ generator a , prove that G2 is also a cyclic group with generator f(a).
i said:
Since f:G1->G2, is an isomorphism and G1 and G2 are cyclic, then G1 and G2 have the same order n , (some n in Z). Therefore since they have the same order, f(a)=a is also cyclic.//

Teacher said I'm trying to show G2 has order n, and that i should use if G1=<a> i have to prove G2=<f(a)>

So since G1=<a> and f:G1->G2 is isomorphic, then a=f(a)
and G2=<a>=<f(a)>

You want the fact that isomorphisms preserve order.

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

i said that, but teacher said it wasn't enough just to say that.

Shadow · **Posted:** Mon, 26 Mar 2012 05:55:25 UTC

DgrayMan wrote:

i said that, but teacher said it wasn't enough just to say that.

Yes, that's just the start, you need to show that $\langle f(a)\rangle\supseteq G_2$ , and you do that by counting elements.

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

G2 is a subset of <f(a)>? now im lost

Shadow · **Posted:** Mon, 26 Mar 2012 06:01:01 UTC

DgrayMan wrote:

G2 is a subset of <f(a)>? now im lost

Equality is two inclusions, it is obvious that $\langle f(a)\subseteq G_2$ , now show the other inclusion and equality holds.

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

So since f is an isomorphism then the order is preserved from G1 to G2, otherwise it would not be an isomorphism right? (I don't see why that isn't enough), But continuing, We know G1=<a> so G1 is a subset of <a> and <a> is a subset of G1 . And somehow its obvious that <f(a)> is a subset of G2, and then we show G2 is a subset of <f(a)> since?...

Shadow · **Posted:** Mon, 26 Mar 2012 17:10:55 UTC

DgrayMan wrote:

So since f is an isomorphism then the order is preserved from G1 to G2, otherwise it would not be an isomorphism right? (I don't see why that isn't enough), But continuing, We know G1=<a> so G1 is a subset of <a> and <a> is a subset of G1 . And somehow its obvious that <f(a)> is a subset of G2, and then we show G2 is a subset of <f(a)> since?...

You've just shown that your function is bijective, not that f(a)

is a generator for it, these things are completely independent.

G_2

is the codomain of

, by the definition of a function and the fact that G_2

is closed under its group operation, $\langle f(a)\rangle\subseteq G_2$ .

What I meant about order preserving is not the order of the group, but the order of

.

S.O.S. Mathematics CyberBoard

Elements whch correspond under an isomorphism.

Who is online