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Artusartos
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I just want to check if there is anything wrong with my understanding...
Let's say we have a group of order 42 that contains [itex]Z_6[/itex]. Since the group of units of [itex]Z_6[/itex] has order (3-1)(2-1), it means that we have 2 elements of order 6 in G, right? In other words, for any cyclic subgroup of order n, we just calculate the group of units to see how many elements of order n we have. Is that correct? Also, if we let [itex]P=Z_6[/itex], we can have at most 7 cyclic subgroups of order 6, since 7=42/|N(P)| (where N(P) is the normalizer of P), if N(P)=6 (which is the smallest it can be, since it has to contain [itex]Z_6[/itex]). I'm just wondering if my understanding is correct.
Thanks in advance
Let's say we have a group of order 42 that contains [itex]Z_6[/itex]. Since the group of units of [itex]Z_6[/itex] has order (3-1)(2-1), it means that we have 2 elements of order 6 in G, right? In other words, for any cyclic subgroup of order n, we just calculate the group of units to see how many elements of order n we have. Is that correct? Also, if we let [itex]P=Z_6[/itex], we can have at most 7 cyclic subgroups of order 6, since 7=42/|N(P)| (where N(P) is the normalizer of P), if N(P)=6 (which is the smallest it can be, since it has to contain [itex]Z_6[/itex]). I'm just wondering if my understanding is correct.
Thanks in advance