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AdrianZ
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Is my proof correct? (abstract algebra - groups)
If G is a finite group of even order, show that there must be an element a≠e such that a=a-1
I believe my proof is a bit odd and unusual, I'd appreciate it if someone else checks it and suggests a more convenient argument for this problem.
well, since G is a finite group of even order, let's assume |G|=2k. since G is finite, we can assume G looks like this: G=\{e,a,a^{-1},b,b^{-1},ab,(ab)^{-1},...\}
But if we relabel all elements, we can show G in the form: G=\{e,g_1,g_1^{-1},...,g_k,g_k^{-1}\}, let's call this new representation of G as G' and notice that G=G'. if we exclude e, we have |G-{e}|=2k-1. the number of g_i's in G' is k, so if all their respective g_i^{-1}'s were distinct, G'-{e} would have 2k elements, but that would be impossible because G and G' were the same set! so that would mean that not all g_i's and g_i^{-1} are distinct, so there exists a g_i for which we have: g_i=g_i^{-1} Q.E.D
Homework Statement
If G is a finite group of even order, show that there must be an element a≠e such that a=a-1
I believe my proof is a bit odd and unusual, I'd appreciate it if someone else checks it and suggests a more convenient argument for this problem.
The Attempt at a Solution
well, since G is a finite group of even order, let's assume |G|=2k. since G is finite, we can assume G looks like this: G=\{e,a,a^{-1},b,b^{-1},ab,(ab)^{-1},...\}
But if we relabel all elements, we can show G in the form: G=\{e,g_1,g_1^{-1},...,g_k,g_k^{-1}\}, let's call this new representation of G as G' and notice that G=G'. if we exclude e, we have |G-{e}|=2k-1. the number of g_i's in G' is k, so if all their respective g_i^{-1}'s were distinct, G'-{e} would have 2k elements, but that would be impossible because G and G' were the same set! so that would mean that not all g_i's and g_i^{-1} are distinct, so there exists a g_i for which we have: g_i=g_i^{-1} Q.E.D
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