6. ISOMORPHISMS
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Example. Let G be the group of real numbers under addition and G be
the group of positive numbers under multiplication. Define (x) = 2x 1. is
1–1 and onto but
(1 + 2) = (3) = 4 6= 2 = 1 · 2 = (1) (2),
so
is not an isomorphism.
Can this mapping be adjusted to make it an isomorphism?
Use (x) = 2x.
If (x) = (y), 2x = 2y =) log2 2x = log2 2y =) x = y, so
Given any positive y, (log2 y) = 2log2 y = y, so
is 1–1.
is onto.
Also, for all x, y 2 R, (x + y) = 2x+y = 2x2y = (x) (y).
Thus
is an isomorphism.
Example. Are U (8) and U (12) isomorphic?
Solution.
U (8) = {1, 3, 5, 7} is noncyclic with |3| = |5| = |7| = 2.
U (12) = {1, 5, 7, 11} is noncyclic with |5| = |7| = |11| = 2.
Define
: U (8) ! U (12) by 1 ! (1), 3 ! 5, 5 ! 7, 7 ! 11.
is clearly 1–1 and onto. Regarding operation preservation:
(3 · 5) = (7) = 11 and (3) (5) = 5 · 7 = 11.
(3 · 7) = (5) = 7 and (3) (7) = 5 · 11 = 7.
(5 · 7) = (3) = 5 and (5) (7) = 7 · 11 = 5.
(1 · 3) = (3) = 5 and (1) (3) = 1 · 5 = 5.
etc.
Since both groups are Abelian, we need only check each pair in a single order.
Thus, U (8) ⇡ U (12).
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