If a group has $n$ elements of order $p$ ($p$ is prime) how many subgroups of order $p$ do we get? If the group was cyclic than only one unique subgroup of order $p$, but what if the group is non-cyclic?
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Turn the question around: if $G$ has $m$ subgroups of order $p$, how many elements of order $p$ does it have? If $x$ has order $p$, then so have $x^2,x^3,\dots,x^{p-1}$, and each of them generates the same subgroup of $G$, namely, $\langle x\rangle$. Thus, if $G$ has $m$ subgroups of order $p$, each of them must contain $p-1$ elements of order $p$, and since $p$ is prime, any two of these subgroups have trivial intersection. Thus, $G$ has $m(p-1)$ elements of order $p$. So if $G$ has $n$ elements of order $p$, what is $m$ (in terms of $p$ and $n$)? |
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Is the group itself of order p or are we discussing the order of the elements of this group? Because if the group itself is of order p (i.e. size p) then there can only be two "subgroups." We can have a subgroup of size 1 (the identity group, {e}) and a subgroup of size p which is the same size as the original group. This arises from Lagrange's Theorem: Let G be a finite group and H a subgroup of G. Then the size of the subgroup divides the size of the group. I take it that G can only have subgroups that divide it. |
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