Properties of
Isomorphisms
Thm 6.2: Isomorphisms
acting on elements
•
Suppose ø is an isomorphism from G
onto G'. Then
1. ø takes the identity of G
to the identity of G'
Proof: e'ø(e) = ø(e) since ø(e) is in G'
and ø(e) = ø(e•e) = ø(e)ø(e) since ø is
OP.
So e'ø(e) = ø(e)ø(e)
Cancel ø(e) on the right to give
e' = ø(e).
2. ø(an) = [ø(a)]n, all n in Z
• Proved for n ≥ 0 by induction. Proved
for n < 0 by showing ø(a-1) = [ø(a)]-1
3.a and b commute iff
ø(a) and ø(b)commute.
Proof: (=>) If ab = ba, then
ø(a)ø(b) = ø(ab) = ø(ba) = ø(b)ø(a)
since ø is operation preserving.
(<=) If ø(a)ø(b) = ø(b)ø(a), then
ø(ab) = ø(a)ø(b) = ø(b)ø(a) = ø(ba) by OP
But ø(ab) = ø(ba) implies
ab = ba since ø is one-to-one.
4. G = <a> iff G' = <ø(a)>
Proof: (=>) Let G = <a>. Set H = <ø(a)>
Since G' is closed, H ≤ G'.
Choose any b in G'. Since ø is onto, there is a
g in G with ø(g) = b.
But g = ak for some integer k, so
b = ø(ak) = [ø(a)]k by part 2.
But then b belongs to H, so G' ≤ H.
It follows that G' = H as required.
4b. G = <a> iff G' = <ø(a)>
Proof: (<=) Suppose G' = <ø(a)>. Let H = <a>.
Since G is closed, H ≤ G. Choose any b in G.
Now ø(b) = [ø(a)]k for some k since G' = <ø(a)>.
So ø(b) = ø(ak) by part 2.
Since ø is one-to-one, b = ak.
Hence b is in H.
It follows that G ≤ H.
So G = H as required.
5. |a| = |ø(a)|
Proof:
[ø(a)]n =e'
iff ø(an) = ø(e)
by (2),(1)
iff an = e
since ø is a 1-1 function
It follows that |a| = |ø(a)|.
6. xk =b and xk = ø(b) have
same number of solutions
Proof:
a is a solution of xk = b
<=> ak = b
<=> ø(ak) = ø(b)
<=> [ø(a)]k = ø(b)
<=> ø(a) is a solution of xk = ø(b)
7. |G|=n => G,G' have the
same number of each order
• This follows from (6)