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Suppose $D_8/N$ is a quotient group. Explain why $N$ cannot be the subgroup $\{I, F\}$ where denotes the flip through the vertical.

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What have you tried? When is the quotient of a group by a subgroup a group? – Tobias Kildetoft Dec 7 '12 at 14:47
You would be better off identifying your difficulties with understanding the definitions and asking about those, than asking us to do your homework. – Alex B. Dec 7 '12 at 15:09
up vote 2 down vote accepted

The quotient group $G/N$ of a group $G$ and a normal subgroup $N$ of $G$ is defined for normal subgroups (as the name "$N$" indicates).

A subgroup $N$ is normal if $gng^{-1}$ is in $N$ for all $n$ in $N$ and all $g$ in $G$.

Does this hold for the flip $n=F$ about the vertical axis of the $8$-gon? (picture of stop sign)? Let's put letters $a$ to $h$ on the corners of the stop sign and then consider the rotation mapping $b$ to $a$, $a$ to $h$, etc. (counter clockwise by one) and denote it by $g=R$.

$RFR^{-1}$ is in $N=\{I,F\}$ if we either have $RFR^{-1} = F$ or $RFR^{-1} = I$. Let's see: we start with the corners labeled $a,b,c,d,e,f,g,h$. After applying $R$ they are labelled $b,c,d,e,f,g,h$. After applying $F$ to this we have $c,b,a,h,g,f,e,d$. Then applying $R^{-1}$ leaves us stuck with $d,c,b,a,h,g,f,e,d$ which is neither $F$ nor $I$. Too bad.

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thanks, but what's picture on – qwerty89 Dec 7 '12 at 15:55
@Downvote Instead of just down voting pointing out the reason why you're down voting would be more useful. – Rudy the Reindeer Dec 7 '12 at 16:11

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