Order Equal to 380

From the factor tables, $380 = 2^{2} \times 5 \times 19$. From Sylow's Theorem, Sylow p-subgroups exist for p = 2, p = 5, p = 19. Therefore there exist at least one subgroup of order 4, one of order 5, and one of order 19. Using the third part of Sylow's Theorem we can form a table of the possible number of cosets for each Sylow p-subgroup. This value is denoted as n_p. Remember, that n_p = 1 mod p and also that $n_{p} \mid m$. We would like to force at least one of these Sylow p-subgroups to be unique and thus normal in the group.

$$\begin{tabular}{\vert c\vert c\vert l\vert} \hline prime & Sy... ...ine 5 & 5 & 1 \\ \hline 19 & 19 & 1,20 \\ \hline \end{tabular}$$

Seeing that n₅ = 1, meaning that the number of Syl₅'s is one, all 5-groups are contained in a Sylow 5-subgroup and so a Syl₅subgroup is normal. Therefore, no group of order 380 can be simple.


This same technique can be used to eliminate many other orders in our range such as; 220, 231, 275, 297, 308, 312, 385.