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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: [itex]G=\{e,a,a^{-1},b,b^{-1},ab,(ab)^{-1},...\}[/itex]
But if we relabel all elements, we can show G in the form: [itex]G=\{e,g_1,g_1^{-1},...,g_k,g_k^{-1}\}[/itex], 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 [itex]g_i[/itex]'s in G' is k, so if all their respective [itex]g_i^{-1}[/itex]'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 [itex]g_i[/itex]'s and [itex]g_i^{-1}[/itex] are distinct, so there exists a [itex]g_i[/itex] for which we have: [itex]g_i[/itex]=[itex]g_i^{-1}[/itex] 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: [itex]G=\{e,a,a^{-1},b,b^{-1},ab,(ab)^{-1},...\}[/itex]
But if we relabel all elements, we can show G in the form: [itex]G=\{e,g_1,g_1^{-1},...,g_k,g_k^{-1}\}[/itex], 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 [itex]g_i[/itex]'s in G' is k, so if all their respective [itex]g_i^{-1}[/itex]'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 [itex]g_i[/itex]'s and [itex]g_i^{-1}[/itex] are distinct, so there exists a [itex]g_i[/itex] for which we have: [itex]g_i[/itex]=[itex]g_i^{-1}[/itex] Q.E.D
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