External direct product- homomorphism

  • #1
Theorem 8.3 in Gallian's Contemp. Abst Alg
says with (s, t) = 1 the group U(st) is isomorphic
to the external direct product of U(s) and U(t)
that is, to U(s) (+) U(t)

U(n) is the group of positive
integers less than n and relatively prime to n with the
group operation multiplication mod n

Defining a map f from U(st) -> U(s)(+) U(t) as f : x -> (x mod s, x mod t)

I am confused on how to proceed to show that f is HOMOMORPHISM. Please see my working below....

LET a, b belong to U( st) then

f ( a x b mod st )=f ( ab mod st)
= ( ( ab mod st ) mod s , ( ab mod st ) mod t))

Now, since ab mod st = ab mod s = ab mod t and (ab mod s ) mod s = ab mod s

= ( ab mod s, ab mod t )
= ( a mods, a mod t) (b mod s , b mod t)
= f ( a ) f (b)
 
Last edited:

Answers and Replies

  • #2
22,089
3,286
That seems correct.

Note however that notations like ( ab mod st ) mod s don't make much sense.
You can write directly f ( ab mod st) = ( ab mod s , ab mod t)
 
  • #3
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,916
19
Note however that notations like ( ab mod st ) mod s don't make much sense.
Actually, it is not unusual to use
x mod y​
to refer to the function that selects a canonical representative of the equivalence class of x modulo y. (commonly the one in the interval [0, y-1])

Also, I don't think it especially surprising to use the postfix operator "- mod f" to refer to the homomorphism G --> G/f. (where G is some appropriate sort of algebraic structure like Abelian group or Ring)
 
  • #4
That seems correct.

Note however that notations like ( ab mod st ) mod s don't make much sense.
You can write directly f ( ab mod st) = ( ab mod s , ab mod t)
Why does my idea fail if take this example

let s= 3 , t = 5. then a = 7 , b= 8 belong to U (st)= U(15)

f ( 7 x 8 mod 15) = ( ( 7 x 8 mod 15 )mod 5, (7 x 8 mod 15) mod 3) = (4, 0).

However , if I write

f ( 7 x 8 mod 15) = ( 56 mod 5, 56 mod 3) = (1, 2).

which is equal to

f ( 7) f (8) = ( 7 mod 5, 7 mod 3)( 8 mod 5, 8 mod 3)

= (2, 1) ( 3, 2)
= ( 1,2)
 
Last edited:
  • #5
22,089
3,286
f ( 7 x 8 mod 15) = ( ( 7 x 8 mod 15 )mod 5, (7 x 8 mod 15) mod 3) = (4, 0).
I don't see how you got this...
 
  • #6
I don't see how you got this...
f ( 7 x 8 mod 15) = ( ( 7 x 8 mod 15 )mod 5, (7 x 8 mod 15) mod 3)
= ( 9 mod 5 , 9 mo d 3)
= ( 4, 0)
 
  • #7
22,089
3,286
7x8=11 mod 15, not 9...
 
  • #8
7x8=11 mod 15, not 9...
Thanks, then it seems my initial reasoning was correct, that is,


( ( ab mod st ) mod s , ( ab mod st ) mod t)) = ( ab mod s, ab mod t ).
 

Related Threads on External direct product- homomorphism

  • Last Post
Replies
3
Views
4K
Replies
2
Views
918
Replies
1
Views
706
Replies
4
Views
717
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
3K
Top