- #1
mehtamonica
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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)
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)
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