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Samuelb88
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Lemma. A group [itex]G[/itex] of order 6 can have only one element of order 3.
Pf. Suppose [itex]G[/itex] has two elements of order 3. Call these elements [itex]x[/itex] and [itex]y[/itex]. Let [itex]H[/itex] and [itex]K[/itex] be the subgroups generated by [itex]x[/itex] and [itex]y[/itex] resp. Then [itex]H \cap K = \{ e \}[/itex] and therefore [itex]G[/itex] can have only one subgroup of order 3.
I'm reading over my notes from class and I'm confused on the reasoning here. Why does [itex]H \cap K = \{ e \}[/itex] imply that [itex]G[/itex] can have only one element of order 3?
Pf. Suppose [itex]G[/itex] has two elements of order 3. Call these elements [itex]x[/itex] and [itex]y[/itex]. Let [itex]H[/itex] and [itex]K[/itex] be the subgroups generated by [itex]x[/itex] and [itex]y[/itex] resp. Then [itex]H \cap K = \{ e \}[/itex] and therefore [itex]G[/itex] can have only one subgroup of order 3.
I'm reading over my notes from class and I'm confused on the reasoning here. Why does [itex]H \cap K = \{ e \}[/itex] imply that [itex]G[/itex] can have only one element of order 3?