Show G is Abelian: Let G be a Finite Group w/ I

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SUMMARY

The discussion centers on proving that a finite group G is abelian if the cardinality of the subset of involutions I, defined as I = {g in G: g^2 = e} \ {e}, satisfies card(I) ≥ (3/4)card(G). Participants highlight that elements in I have order 2 and that all elements in the union of I and the identity element e commute. The solution involves using Lagrange's theorem and analyzing the centralizer subgroup of elements in I to demonstrate commutativity throughout G.

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  • Understanding of group theory concepts, specifically finite groups and abelian groups.
  • Familiarity with the definition and properties of involutions in group theory.
  • Knowledge of Lagrange's theorem and its implications for subgroup orders.
  • Basic understanding of centralizer subgroups and their role in group structure.
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  • Study the proof of Lagrange's theorem and its applications in group theory.
  • Explore the properties of centralizer subgroups and their significance in proving group commutativity.
  • Investigate the characteristics of involutions and their impact on group structure.
  • Learn about the classification of finite groups and the conditions under which they are abelian.
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This discussion is beneficial for students of abstract algebra, particularly those studying group theory, as well as mathematicians interested in the properties of finite groups and their structures.

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Homework Statement



Let G be a finite group and let I ={g in G: g^2 = e} \ {e} be its subset of involutions. Show that G is abelian if card(I) => (3/4)card(G).

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The Attempt at a Solution

I don't really know how to proceed with this problem and to make use of 3/4. I know that the set I is precisely the elements of order 2 in the group G and all elements within I U {e} commute; but I just don't know how to show the whole group commutes. Any help on how to proceed with the problem is highly appreciated. Thanks.
 
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I'll give you a hint while I'm trying to figure this one out myself. You don't know that I commutes, do you? It's perfectly possible that a^2=e and b^2=e and ab is not equal to ba, isn't it? If you HAVE figured out how to prove {e} U I is commutative, then you would be done. That makes it a subgroup. Use Lagrange's theorem.
 
For future reference, I think I finally got this one. Consider an element a in I. Now look at the |G| equations aG=G. Show if ab=c and b and c are also in I, then b commutes with a. Count how many equations in aG=G are of that form and draw a conclusion about the size of the centralizer subgroup of a. The rest is an easy exercise. That was bothering me.
 

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