# Why can $N$ not be the subgroup $\{I, F\}$?

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

-
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

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.