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I have to show that the number of subgroups of $\mathbb{Z}_{5}\times\mathbb{Z}_{5}$ (other than identity and itself) is six, but I am a bit confused. please help!

P.S. can someone also tell if there is some method to determine the subgroups of direct products of cyclic groups?

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The subgroups you are looking for must have order 5. In particular, they are cyclic. The elements $(0,1),(1,0),(1,1),(1,2),(1,3),(1,4)$ each generate one of these subgroups.

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