Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 5.1 Exercise 5.1.10
Solution: We saw previously (by counting) that $E$ has precisely $p+1$ distinct subgroups of order $p$. Thus it suffices to show that these subgroups are mutually distinct.
Suppose $\langle (x^a,y^b) \rangle = \langle (x^c,y^d) \rangle$. Then $(x^c,y^d) = (x^a,y^b)^k$ for some $k$, hence $$(x^{ak},y^{bk}) = (x^c,y^d).$$ Thus $x^{ak} = x^c$, so that $ak \equiv c \pmod p$. Likewise, $bk \equiv d \pmod p$.
Suppose $a \equiv 1$. Then $c \not\equiv 0$ since otherwise the first coordinate of each element of $\langle (x^c,y^d) \rangle$ is 1, a contradiction since $a \neq 1$. Thus none of the first $p$ subgroups above is equal to the last. If (without loss of generality, considering the generators chosen above) $c \equiv 1$, then $k \equiv 1$, and we have $b \equiv d \pmod p$. Thus the first $p$ subgroups in the list above are mutually distinct. Hence the subgroups given above exhaust the distinct order $p$-subgroups of $E$.
