- #26

- 1,086

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Is that because there is only one subgroup of order 126 in [itex]\mathbb{Z}_{252}[/itex]? How would we know that, though? I mean, I guess both the subgroup of order 126 (which would be cyclic) and [itex]\mathbb{Z}_{252}[/itex] have the same number of elements of order 126, so you could not have two such subgroups of that order, but how do you show a group has a certain number of elements of a particular order? With all the other homework to do, I'm just tuning in and out of this, so I'm always really puzzled when I get back to it... looking at the decomposition [itex]G\cong (\mathbb{Z}_5\times \mathbb{Z}_5)\times \mathbb{Z}_{252}[/itex] yields that there is exactly one subgroup of G of order 630 corresponding to each subgroup of [itex]\mathbb{Z}_5\times \mathbb{Z}_5[/itex] of order 5 ...