Isomorphism of a quotient group

Asked Feb 07 '16 at 18:12

Active Feb 07 '16 at 18:12

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I have that $G=S_4$ and $N = \{1, (12)(34), (13)(24), (14)(23)\}$, and thus far I have shown that N is a normal subgroup of G. I'm trying to figure out what group $G/N$ will be isomorphic to, but I can't figure it out. Any ideas on what the group is, and how I'd proceed proving that is the case?

I think the subgroup will be of order 2, but past that I'm not sure.

asked Feb 07 '16 at 18:12

Ian Baker

If $G$ and $N$ are finite, then $|G / N| = |G| / |N|$, so in this case $|G / N| = 6$, and up to isomorphism there are only two groups of this order. – Travis Willse Feb 07 '16 at 18:16
Okay, these are $S_3$ and $C_6$? - but how would I go about proving that its not isomorphic to the cyclic group? – Ian Baker Feb 07 '16 at 19:45
See the question linked at the top of the question body. – Travis Willse Feb 07 '16 at 20:11
I don't quite follow it - I simply list the cosets, then check a single element of each doesn't have order 6 and that will be sufficient? – Ian Baker Feb 07 '16 at 20:15
Yes---if we instead had $G / H \cong C_6$, there would be an element of order $6$ in $G / H$. – Travis Willse Feb 07 '16 at 20:16

Isomorphism of a quotient group

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