If G is a group of order 35, what are the possible orders for the elements of the group?

By Lagrange the possible orders for G's elements are 1 5, 7, and 35.  

If G has no elements of order 5 then all elements have either order 7.  Let H = (h) the subgroup generated by h, an element of order 7.  Since H has only 7 elements it does not exhaust G, so there must be an element g in G that's not in H.  

Then g has order 7 too and the cosets H, gH, g^2H, ...g^6H must all be distinct.  Otherwise (g^m)(h^i) = (g^n)(h^j), ie, g^(m-n)=h^(j-i).  Selecting b such that (m-n)b = 1 mod 7 you have g = g^(m-n)b = h^(j-i)b, ie g is in H - contradiction.

Notice that there are 49, not 35 elements in the union of the cosets H, gH, g^2H, ...g^6H - a contradiction.  So G must have an element of order 5.

suppose every non-identity element of G was of order 7.

we get 6 non-identity elements in for every such g of order 7.

since two distinct subgroups and only intersect in the identity

(because 7 is prime), the total number of elements of G must therefore be:

|G| = 1 + 6k, for some integer k.

this implies 34 is divisible by 6, which provides our contradiction.