I have a subgroup $N$ of $S_4$, where $ N = [1, (1,2)(3,4), (1,3)(2,4), (1,4)(2,3)] $ I need to explain whether quotient group $G/N$ is isomoprhic to either $C_6$ or $D_6$ (no proof required, just an explanation to why its isomorphic to one and not the other). Now i know its $D_6$ as N doesn't have a generator element and is not cyclic but this is a weak explanation, I can't spot any other differences.
Asked
Active
Viewed 135 times
2
-
what group is $G$? – mjb4 Feb 02 '14 at 16:26
-
Sorry, G is $S_4$ – user65972 Feb 02 '14 at 16:31
-
ah okay. Anyway you got your answer already! $C_6$ is cyclic wheres $D_6$ is not so $C_6\not\simeq D_6$. – mjb4 Feb 02 '14 at 16:33
-
That's not a weak answer, it's a perfectly valid answer. – hmmmm Feb 02 '14 at 16:42
-
2But the fact that $N$ is not cyclic does not imply that $G/N$ is not cyclic, so you won't get many marks for that explanation! – Derek Holt Feb 02 '14 at 17:48
-
possible duplicate of An epimorphism from $S_{4}$ to $S_{3}$ having the kernel isomorphic to Klein four-group – Jyrki Lahtonen Feb 02 '14 at 19:12
3 Answers
3
Hint: Suppose there was a surjective homomorphism $\varphi\colon S_4\to C_6$ with $H$ as the kernel. What would the preimage of a generator of $C_6$ have to satisfy?
Dan Rust
- 30,108
0
Suppose $G/N\simeq C_6$. Then $G/N$ contains a unique, and therefore normal, subgroup of order 2. Let $H \le G$ be the pre-image of this group under the natural homomorphism $G \to G/N$. Then $H$ is a normal Sylow-2-subgroup of $G$ which contradicts the structure of $S_4$.
jpvee
- 3,563