How do we find an example of nonabelian group for which all proper subgroups are normal?? It's one of the questions on my study-guide sheet. Thank you
Asked
Active
Viewed 6,344 times
13
-
@KReiser You are correct. I was thinking "non-cyclic" vs. "non-abelian". Thanks for pointing that out! – amWhy Nov 13 '12 at 01:02
-
1Do you know what the quaternion group is? – Miha Habič Nov 13 '12 at 01:31
-
3http://planetmath.org/HamiltonianGroup.html – Makoto Kato Nov 13 '12 at 01:58
-
Can we give an example other than the quaternions? – Saaqib Mahmood Apr 06 '14 at 07:14
1 Answers
13
Let $Q = \{1, -1, i, -i, j, -j, k, -k\}$ be the quaternion group. Let $Z$ be a subgroup of order $2$. Since $-1$ is the only element of order $2$, $Z = \{1, -1\}$. Since it is the only subgroup of order $2$, it is normal. Let $H$ be a subgroup of order $4$. Since $(Q : H) = 2, Q = H \cup aH = H \cup Ha$ for every $a \in Q - H$. Hence $aH = Ha$. Hence $H$ is normal.
Makoto Kato
- 42,602
-
Why the fact that $Z$ is the only subgroup of order $2$ implies it's normal? – user42912 Mar 31 '14 at 18:27
-
2@user42912 Because $aZa^{-1}$ is a subgroup of order $2$ for every $a \in Q$. – Makoto Kato Mar 31 '14 at 18:32