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!

It's going to have fewer than $7$, but proving that requires some tools it doesn't sound like you have. 
Two distinct subgroups of order $5$ intersect only in the identity (by Lagrange's theorem). This means no two will share any nonidentity elements. Thus if there are $m$ subgroups of order $5$, you can calculate exactly how many elements are in their union. This can't be more than $30$, the order of the group.
