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The derived subgroup is the subgroup generated by the set of all commutators of a group $G$.

I always used to forget that "generated by" part. Soon I will be teaching a group theory course and wish to prevent students from making the same mistake.

Is there an easy example, presentable to beginning group theory students, of a group in which the set of commutators is proper in the derived subgroup?

I am aware of this question, but the students I will be talking to will be below the level of wreath products, so the paper linked there isn't of any use. Is there perhaps an example in the infinite groups which would be easier to understand?

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I think this is true of the free group on two generators. –  user29743 Dec 19 '12 at 19:23
    
There are (many) infinite groups where this is true; proving it so is an entirely different question... –  user641 Dec 19 '12 at 19:49
    
See J.J.Rotman's book of An Introduction to the theory of groups, page 34. I hope it helps you. Nice question. –  B. S. Dec 19 '12 at 19:53
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Related - math.stackexchange.com/questions/7811/… –  anon Dec 19 '12 at 20:02
    
wrt anon's link, Arturo's answer tells us that the smallest such group has order 96. –  user1729 Dec 20 '12 at 16:12
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1 Answer

up vote 3 down vote accepted

This publication On Commutator Groups, by Kappe and Morse is great to read. Among many, it contains the following theorem:
If $G$ is a finite group with $|G : Z(G)|^2 < |G'|$, then there are elements in $G'$ that are not commutators.
With the help of this one can construct a large family of groups of nilpotency class 2 with the property that the set of commutators is not equal to the commutator subgroup.

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