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moont14263
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Let G be a finite group. Suppose that every element of order 2 of G has a complement in G, then G has no element of order 4.
Proof. Let x be an element of G of order 4. By hypothesis, G=<x^{2}> K and < x^{2}> [itex]\cap[/itex]K=1 for some subgroup K of G. Clearly, G=< x> K and < x>[itex]\cap[/itex] K=1$, but |G|=|< x^{2}>||K|<|< x >||K|=|G|, a contradiction. Therefore G has no element of order 4.
Is above true? Thanks in advance.
Proof. Let x be an element of G of order 4. By hypothesis, G=<x^{2}> K and < x^{2}> [itex]\cap[/itex]K=1 for some subgroup K of G. Clearly, G=< x> K and < x>[itex]\cap[/itex] K=1$, but |G|=|< x^{2}>||K|<|< x >||K|=|G|, a contradiction. Therefore G has no element of order 4.
Is above true? Thanks in advance.