Example 2: Show that a finite group is not free on any set. Solution: Let F be a finite group. Suppose F is free on a subset S={S₁, S₂, ..., Sn}. Consider a:S-Z:a(s) = r. Then, by the universal mapping property, there will be a unique group homomorphism ã: F→ Z:ã(s,) = r. Note, that a(ss) =ã(s) +ã(s) = m+k for m, k = 1, ..., n. Now, se F, so that se, where o(F) = t and m = 1, . m n. So, a(s)= a(e) = 0, that is, ta(s) = 0, that is, tm = 0, which is not possible since t=0 m +0
Example 2: Show that a finite group is not free on any set. Solution: Let F be a finite group. Suppose F is free on a subset S={S₁, S₂, ..., Sn}. Consider a:S-Z:a(s) = r. Then, by the universal mapping property, there will be a unique group homomorphism ã: F→ Z:ã(s,) = r. Note, that a(ss) =ã(s) +ã(s) = m+k for m, k = 1, ..., n. Now, se F, so that se, where o(F) = t and m = 1, . m n. So, a(s)= a(e) = 0, that is, ta(s) = 0, that is, tm = 0, which is not possible since t=0 m +0
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:Example 2: Show that a finite group is not free on any set.
Solution: Let F be a finite group. Suppose F is free on a subset
S = {S₁, S₂, ..., Sn}.
Consider a:S-Z:α(s,)=r.
Then, by the universal mapping property, there will be a unique group
homomorphism ã: F→ Z:ã(s) = r.
Note, that a(ss)= a(s) +ã(s) = m+k for m, k = 1, ..., n.
m
Now, se F, so that se, where o(F) = t and m = 1, ..., n.
m
So, a(s)= a(e) = 0, that is, ta(s) = 0, that is, tm = 0, which is not possible
since t#0, m #0.
We reach a contradiction. Hence F is not free.
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