Im reading Dummit and foote page 145 where the book show $G$ is simple whenever $|G|=60$ and $n_5>1$
The book did it with a contradiction by assuming it is simple. Then deduced right away that $n_5=6$ but I do not see how this follows, I think $n_5=11$ can also hold here because $11*4=44$ in this case $n_3=3$ and $n_2=3$ We have $44+6+9=59$ So we have 59 distinct elements in total add up to 60 with the identity. I dont see how $n_5$ must be 6.
Sylow's third theorem not only says that $n_{5}$ is congruent to 1 mod 5, it also says that $n_{5}$ divides the largest factor of $|G|$ that is not divisible by 5. (In Dummit and Foote, this is phrased by saying that if $|G| = p^{\alpha}m$, where $p$ does not divide $m$, then $n_{p}$ divides $m$.) In this case, we see that $n_{5}$ must divide 12, ruling out $n_{5} = 11$.