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PostPosted: Tue, 28 Feb 2012 04:54:45 UTC 
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G1 and G2 are groups and f: G1-> G2 is an isomorphism.

if e1 is the neutral element of G1 and e2 is the neutral element of G2, how do you prove f(e1)=e2 ??

I know there is only one identity element in each group. and i suppose since an isomorphism is bijective then e1=e2.


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PostPosted: Tue, 28 Feb 2012 04:57:35 UTC 
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DgrayMan wrote:
G1 and G2 are groups and f: G1-> G2 is an isomorphism.

if e1 is the neutral element of G1 and e2 is the neutral element of G2, how do you prove f(e1)=e2 ??

I know there is only one identity element in each group. and i suppose since an isomorphism is bijective then e1=e2.


No, that's ridiculous. Instead let g_2\in G_2, then since f is a bijection, \exists g_1\in G_1 such that f(g_1)=g_2

Then note that g_2f(e_1)=f(g_1)f(e_1)=f(g_1e_1)=f(e_1g_2)=f(e_1)f(g_1)=f(e_1)g_2=g_2, the last equality following from the fact that f(e_1g_1)=f(g_1)=g_2.

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PostPosted: Tue, 28 Feb 2012 05:09:03 UTC 
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what about e2? i don't really understand how f(e1)=e2


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PostPosted: Tue, 28 Feb 2012 05:16:35 UTC 
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DgrayMan wrote:
what about e2? i don't really understand how f(e1)=e2


By definition the neutral element fixes every element of a group when multiplied on the left and right, you have shown that f(e_1) does that, so it is the neutral element.

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PostPosted: Tue, 28 Feb 2012 05:40:44 UTC 
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Shadow wrote:
DgrayMan wrote:
G1 and G2 are groups and f: G1-> G2 is an isomorphism.

if e1 is the neutral element of G1 and e2 is the neutral element of G2, how do you prove f(e1)=e2 ??

I know there is only one identity element in each group. and i suppose since an isomorphism is bijective then e1=e2.


No, that's ridiculous. Instead let g_2\in G_2, then since f is a bijection, \exists g_1\in G_1 such that f(g_1)=g_2

Then note that g_2f(e_1)=f(g_1)f(e_1)=f(g_1e_1)=f(e_1g_2)=f(e_1)f(g_1)=f(e_1)g_2=g_2, the last equality following from the fact that f(e_1g_1)=f(g_1)=g_2.


Or, slightly differently,
f(e_1)f(e_1)=f(e_1e_1)=f(e_1)
and the only idempotent element in G_2 is ...

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Tue, 28 Feb 2012 05:51:24 UTC 
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Ah i get it. Heres another problem with the same G1 and G2.
If G1 is a cyclic group with generator a, how do i prove G2 is also a cyclic group of generator f(a)?

So i can visualize it in my head that a=f(a) if its isomorphic, and if a is a generator of G1 then f(a) is a generator G2. But how do i show this?


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PostPosted: Tue, 28 Feb 2012 06:21:33 UTC 
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DgrayMan wrote:
Ah i get it. Heres another problem with the same G1 and G2.
If G1 is a cyclic group with generator a, how do i prove G2 is also a cyclic group of generator f(a)?

So i can visualize it in my head that a=f(a) if its isomorphic, and if a is a generator of G1 then f(a) is a generator G2. But how do i show this?


This is obvious, just use the definition of the order of a group. You should think about these before you ask here.

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PostPosted: Tue, 28 Feb 2012 15:13:58 UTC 
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i wish i was a math genius like you guys, but im not. i struggle with the obvious. It's kind of like being asked a question such as "how old you are?" in a foreign language. Of course i know the answer, but i don't understand what it is they're asking.


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PostPosted: Tue, 28 Feb 2012 15:37:41 UTC 
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DgrayMan wrote:
i wish i was a math genius like you guys, but im not. i struggle with the obvious. It's kind of like being asked a question such as "how old you are?" in a foreign language. Of course i know the answer, but i don't understand what it is they're asking.


I'm not saying you should expect to get them ALL right away, but this one doesn't take a genius, the definition of the order of an element and the order of a group and of a cyclic group are all you need, no theorems, no lemmata, no nothing else, just the definitions and understanding what the question is asking you. This problem is not an exercise in mathematics, it is an exercise in understanding what you are being asked. Think about it, and if you're still having trouble feel free to post your thoughts on the matter/what you came up with and we'll be glad to take a look and help some more.

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PostPosted: Sun, 4 Mar 2012 01:06:29 UTC 
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[quote="DgrayMan"]Ah i get it. Heres another problem with the same G1 and G2.
If G1 is a cyclic group with generator a, how do i prove G2 is also a cyclic group of generator f(a)?
/quote]
No wonder i didn't know this answer, this book goes over order of group elements in the NEXT chapter. Sorry guys.

So if a is a generator of G1 and it has n elements since it is finite then the order of a is n since e=a^n, i believe is the definition.

So since they both have the same neutral element does this prove f(a^2)=e so a=f(a) ? ? Probably not....


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PostPosted: Sun, 4 Mar 2012 01:23:37 UTC 
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DgrayMan wrote:
DgrayMan wrote:
Ah i get it. Heres another problem with the same G1 and G2.
If G1 is a cyclic group with generator a, how do i prove G2 is also a cyclic group of generator f(a)?
/quote]
No wonder i didn't know this answer, this book goes over order of group elements in the NEXT chapter. Sorry guys.

So if a is a generator of G1 and it has n elements since it is finite then the order of a is n since e=a^n, i believe is the definition.

So since they both have the same neutral element does this prove f(a^2)=e so a=f(a) ? ? Probably not....


Use the definition of cyclic and the fact that isomorphisms preserve order.

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PostPosted: Mon, 5 Mar 2012 01:38:59 UTC 
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Can you show me an example?


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PostPosted: Mon, 5 Mar 2012 03:10:26 UTC 
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Consider the sets:

A=\{\pm 1\} and B=\{[0],[1]\}.

Let an operation *_A be defined on A where a_1*_Aa_2=a_1a_2 where the multiplication on the right is normal multiplication of real numbers, and on B define *_B by b_1*_Bb_2=[b_1+b_2], where as usual [] denotes the equivalence class of a number modulo 2.

Then the map defined by f(-1)=1 as a map from A\to B is an isomorphism because it maps an element of order 2 in A to an element of order 2 in B, and since B has order 2 as a group, that implies it is cyclic.

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PostPosted: Mon, 5 Mar 2012 04:47:04 UTC 
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If I'm trying to find an isomorphism between two groups. Say i make a group table and im able to find the neutral elements in each row and column first. If both tables match up does this mean they have the same order and therefore an isomorphism? So i guess basically i'm asking, if the neutral elements in both tables match up is it an isomorphism?


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PostPosted: Mon, 5 Mar 2012 17:03:03 UTC 
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DgrayMan wrote:
If I'm trying to find an isomorphism between two groups. Say i make a group table and im able to find the neutral elements in each row and column first. If both tables match up does this mean they have the same order and therefore an isomorphism? So i guess basically i'm asking, if the neutral elements in both tables match up is it an isomorphism?


No, tables have inherent orders to them, there is nothing about the neutral element that can determine a group. All groups have a neutral element, so it is the least useful invariant you can possibly have. You need to find a map (function) which is an isomorphism or use a theorem which says one exists under whatever conditions you have satisfied. In your case, I've explicitly described the map as well as an alternative method using orders and the definition of a cyclic group. Have you tried either?

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