Question about sylow p-subgroup. Question: Find the number of sylow 5-subgroup of group of order 60
My attempt: 
  O(G)=60=2^2•3•5

As,5|60 and 5^2 does not divide 60 so, by sylow first theorem G must have at least one sylow 5-subgroup of order 5.
Now by sylow 3rd theorem number of sylow 5-subgroup say n_5 satisfies,
n_5 = 1+5k and n_5 | 60
Both of above holds for k=0,1.
But in text book they take k= 1 and so n_5 = 6.
Why they take k=1? and not take k=0?
$n_5$ satisfies two conditions:

1) $n_5\mid |G|/5=12$ which implies $n_5\in \{1,2,3,4,6, 12\}$

2) $n_5\equiv 1 {\pmod 5}$

The two conditions are satisfied only if $n_5=6$.
$n_5$ satisfies two conditions:

1) $n_5\mid |G|/5=12$ which implies $n_5\in \{1,2,3,4,6, 12\}$

2) $n_5\equiv 1 {\pmod 5}$

The two conditions are satisfied only if $n_5=6$.