A group of order 30 has at most 7 subgroups of order 5 
Show that a group of order 30 has at most 7 subgroups of order 5.

This should be a basic question (from an introductory algebra course), but I got no clue... Please help!

Apply Sylow theorem,
$$30 = 2 \times 3 \times 5.$$ 
No. Of Sylow 5 sub-groups $= 1 \mod 5$ and divides 6, i.e. 1 or 6
So, max 6 subgroups of order 5 is possible.
