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Permissible subgroup orders in a group of order 120, and their indices


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.2 Exercise 3.2.1

Which of the following are permissible orders for subgroups of a group of order 120: 1, 2, 5, 7, 9, 15, 60, 240? For each permissible order give the corresponding index.


Solution: Let $G$ be a group and $N \leq G$ a subgroup.

(1) $|N| = 1$ is possible because $1|120$. If such a subgroup exists, then $[G:N] = 120/1 = 120$.
(2) $|N| = 2$ is possible because $2|120$. If such a subgroup exists, then $[G:N] = 120/2 = 60$.
(3) $|N| = 5$ is possible because $5|120$. If such a subgroup exists, then $[G:N] = 120/5 = 24$.
(4) $|N| = 7$ is not possible because 7 does not divide 120.
(5) $|N| = 9$ is not possible because 9 does not divide 120.
(6) $|N| = 15$ is possible because $15|120$. If such a subgroup exists, then $[G:N] = 120/15 = 8$.
(7) $|N| = 60$ is possible because $60|120$. If such a subgroup exists, then $[G:N] = 120/60 = 2$.
(8) $|N| = 240$ is not possible because $240 > 120$.


Linearity

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