Proving Even Order Group G Has Element a=/e Satisfying a^2=e

Thread starter fk378
Start date Sep 16, 2008
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SUMMARY

The discussion centers on proving that a group G of even order contains an element a ≠ e such that a² = e. Participants clarify that if a = a⁻¹, then a is its own inverse, leading to the conclusion that G must have a subgroup of order 2. The key argument hinges on the fact that if G has an even number of elements, the identity element e pairs with itself, leaving an odd number of non-identity elements. This contradiction implies the existence of at least one element a ≠ e such that a² = e, confirming the original statement.

PREREQUISITES

Understanding of group theory concepts, specifically group order and inverses.
Familiarity with equivalence relations and their properties.
Knowledge of Lagrange's theorem and its implications for subgroup orders.
Basic proof techniques, particularly contradiction proofs.

NEXT STEPS

Study the implications of Lagrange's theorem in group theory.
Learn about equivalence relations and their applications in mathematics.
Explore examples of groups with even and odd orders, such as the Klein four-group.
Practice constructing proofs by contradiction in abstract algebra.

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Mathematics students, particularly those studying abstract algebra, group theory enthusiasts, and anyone interested in understanding the properties of groups with even order.

Sep 20, 2008

HallsofIvy

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fk378 said:

I'm a bit confused by the wording of this statement. Do you mean that since we now have |G|=odd, and then we multiply that number of sets with (2), then the order will actually become even?

If I understand what you are saying here, yes, that's the basic idea. IF G contains no member a, other than the identity e, such that a^-1= a, then we can pair all non-identity members, {a, a^-1}. If we multiply the number of such sets by 2, we get the number of non-identity members which must be even. Since there is one identity, e, the total number of members of G, |G|, must be odd.

Here's an example. Suppose we have a group with 4 members, e, x, y and z where e is the group identity. Suppose further that there is no member, other than e, such that a^-1= a. e is its own inverse. What is the inverse of x? Since x is not its own inverse, it must be either y or z. Suppose it is y: we can pair {x,y}. Now that leaves z. z is not, by hypothesis, its own inverse, but there is nothing else left to pair it with!

It occurs to me that this requires we know that the inverse of z cannot be either x or y because they are "already paired"- theorem: if a^-1= b and c^-1= b, the a= c. Do you see why that is true? Suppose a^-1= b and c^-1= b. Look at a(cb) and (ac)b.