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2. Let ??????:�→� under addition given by ?????? � =5�.
Answer: The identity element of the set of images is 0 under the operation of addition. If 5x
= 0, then x = 0. Thus, Ker(??????)={0}.
COROLLARY:
A group homomorphism ??????:??????→??????′ is one – to – one map if and only if Ker(??????) = {??????}.
In view of this corollary, we can modify our steps in showing that two groups are isomorphic.
To show ??????:??????→??????′ is an isomorphism: (modified version)
1. Show ?????? is a homomorphism.
2. Show Ker(??????) = {??????}.
3. Show ?????? maps G onto G’.
Examples: (On showing isomorphism between G and G’)
A. Let us show that the binary structure <R, +> with operation the usual addition is isomorphic to
the structure <R
+, ∙> where ∙ is the usual multiplication.
Step 1. We have to somehow convert an operation of addition to multiplication. Recall from
�
�+�
= �
�
(�
�
) that the addition of exponents corresponds to multiplication of two quantities.
Thus we try defining ??????:�→�
+
by ?????? � =??????
�
for �∈�. Note that ??????
�
>0 for all �∈� so indeed
?????? � ∈�
+
.
Step 2. If ?????? � =??????(�), then ??????
�
=??????
�
. Taking the natural logarithm, we see that �=�, so ?????? is
indeed 1-1.
Step 3. If ??????∈�
+
, then ln(??????)∈� and ?????? ln?????? =??????
ln ??????
=??????. Thus ?????? is onto R
+.
Step 4.For�,�∈�, we have?????? �+� =??????
�+�
=??????
�
∙??????
�
=?????? � ∙?????? � .
Thus, ?????? is an isomorphism.
B. Let 2�= 2� �∈� , so that 2Z is the set of all even integers, positive, negative and zero. We
claim that <Z, +> is isomorphic to <2Z, +> where + is the usual addition.
Step 1. Define ??????:�→2� by ?????? � =2� for �∈�.
Step 2. If ?????? � =??????(�), then 2�=2� so �=�. Thus ?????? is 1-1.
Step 3. If �∈2�, then n is even so �=2� for �=�/2∈�. Hence ?????? � =2(�/2)=� so ?????? is onto
2Z.
Step 4. For�,�∈�, we have?????? �+� =2 �+� =2�+2�=?????? � +??????(�).
Thus, ?????? is an isomorphism.
C. Determine whether the given map ?????? is an isomorphism of the first binary structure with the
second.
1. <Z, +> with <Z, +> where ?????? � =2� for �∈�.
Answer: ?????? is 1-1, but NOT onto. Thus, ?????? is NOT an isomorphism.
2. <Z, +> with <Z, +> where ?????? � =�+1 for �∈�.
Answer:?????? is 1-1, onto, but NOT operation preserving. Thus, ?????? is NOT an isomorphism.