Prove two groups are isomorphic

Question

I have created the two tables but can not find a one to one correlation between the values in the two tables. I would appreciate it if anyone can point me in the right direction to understand how to solve this.

+4  0   1   2   3  
0   0   1   2   3  
1   1   2   3   4  
2   2   3   0   1  
3   3   4   1   2  

*5  1   2   3   4  
1   1   2   3   4  
2   2   4   1   3  
3   3   1   4   2  
4   4   3   2   1

Thank you!

question 31.3-1

It would be very helpful for you to provide the tables of the group, because we could help find a more specific answer — wjmccann, Jul 22 '19 at 21:25
Ok, to start with, what does $0$ correspond to? and are you familiar with cyclic groups? — J. W. Tanner, Jul 22 '19 at 21:44
based on the table 0 corresponds to 1 in the *5 table. I am not familiar with cyclic groups — J. Doe, Jul 22 '19 at 21:48
correct; do you know what the order of an element is? orders are preserved by isomorphisms — J. W. Tanner, Jul 22 '19 at 21:50
You have $G=({ 0,1,2,3},+_4)$ and $H=({1,2,3,4},\ast_5)$ If they are isomorphic, then exists a bijective homomorphism $f\colon G\to H$ such that:
$$ f(a+_4b)=f(a)\ast_5f(b) $$ — luisfelipe18, Jul 22 '19 at 21:55
the order of +4={0,1,2,3} is (1,2,3,3). Order of *5={1,2,3,4} is (1,2,2,2) — J. Doe, Jul 22 '19 at 22:38
No, the order of $1$ and $3$ in $\Bbb Z_4$ is $4$; the order of $2$ in $\Bbb Z_4$ is $2$; the order of $2$ and $3$ in $\Bbb Z_5^*$ is $4$; so an isomorphism must map $2$ to $4$ — J. W. Tanner, Jul 22 '19 at 22:52
I see your point. now we have two more that must be mapped but they all have order of 4. — J. Doe, Jul 22 '19 at 23:03
yes, in fact you could map $1$ to $2$ or $1$ to $3$ (and $3$ to $3$ or $3$ to $2$, respectively) and have an isomorphism — J. W. Tanner, Jul 22 '19 at 23:06
are you sure the order of Z* of 2 and 3 is 4? because when I do 2^2mod5=4 order is 2 — J. Doe, Jul 22 '19 at 23:43
order in $\Bbb Z_5^*$ of $2$ and $3$ is $4,$ because powers of $2$ are $1,2,4,$ and $3$ and $2^4=16\equiv1,$ and powers of $3$ are $1,3,9\equiv4,$ and $27\equiv2$ and $3^4=81\equiv1$ — J. W. Tanner, Jul 23 '19 at 00:36

J. W. Tanner · Accepted Answer · 2019-07-22T23:07:03.260

The groups are cyclic; that means they are generated by a single element.

A generator of $\Bbb Z_4$ is $1$; i.e., the elements in $\Bbb Z_4$ are $0, 1, 1+1, $ and $1+1+1$.

A generator of $\Bbb Z_5^*$ is $2$; i.e., the elements in $\Bbb Z_5^*$ are $1, 2, 2^2, 2^3\equiv_53. $

The map from $\Bbb Z_4$ to $\Bbb Z_5^*$ that takes $0$ to $1$, $1$ to $2$, $1+1$ to $2\times2$, and $1+1+1$ to $2^3\equiv_5 3$

is an isomorphism.

To prove that from scratch, you would have to verify $16$ statements

(actually only $10$ if you use commutativity).

Prove two groups are isomorphic

1 Answers1

Linked