If a group G has at least 7 elements of order 7 then prove t

If a group G has at least 7 elements of order 7, then prove that G has at least 12 elements of order 7.

Solution

Suppose n = 7. Then j G j= 7 7 = 49. Since 7 is the only positive divisor of 49 between 1
and 49, it is the only possible order of a subgroup other than feg or G. FTCG also tells us
that there is exactly one subgroup of order 7. This is the supposed criteria

In general, if we suppose that n is any positive integer besides 7, we see that G is guaranteed a
subgroup of order n by the FTCG, which means that G will have at least 4 distinct subgroups.
We therefore conclude that the order of G must be 72 = 49.
{ More generally, if 7 is replaced by any prime p under the supposed conditions, the the order of G
must be p^2