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I am asked to show that $\langle r_4 \rangle$ is a normal subgroup of $D_8$, where

$D_8 = \{r_0, r_1, r_2, r_3, r_4, r_5, r_6, r_7, s_0, s_1, s_2, s_3, s_4, s_5, s_6, s_7 \}.$

I am also asked to construct a Cayley table for the quotient group $D_8 / \langle r_4 \rangle$ and see which group this is isomorphic to.

To show that $\langle r_4 \rangle$ is a normal subgroup of $D_8$ I need to show that all the left cosets coincide with all of the right cosets. Here is my work.

$\langle r_4 \rangle = \{ r_0, r_4 \}$

$r_1 \langle r_4 \rangle = \{r_1, r_5 \} = \langle r_4 \rangle r_1$

$r_2 \langle r_4 \rangle = \{r_2, r_6 \} = \langle r_4 \rangle r_2$

$r_3 \langle r_4 \rangle = \{r_3, r_7 \} = \langle r_4 \rangle r_3$

$s_0 \langle r_4 \rangle = \{s_0, s_4 \} = \langle r_4 \rangle s_0$

$s_1 \langle r_4 \rangle = \{s_1, s_5 \} = \langle r_4 \rangle s_1$

$s_2 \langle r_4 \rangle = \{s_2, s_6 \} = \langle r_4 \rangle s_2$

$s_3 \langle r_4 \rangle = \{s_3, s_7 \} = \langle r_4 \rangle s_3$

Hence $\langle r_4 \rangle \unlhd D_8.$

Here is the Caley table for the quotient group $D_8 / \langle r_4 \rangle$, where $N = \langle r_4 \rangle$.

$r_i r_j = r_k$ & $r_i s_j = s_k$ where $k = (i + j) mod 8$ while

$s_i r_j = s_l$ & $s_i s_j = r_l$ where $l = (i - j) mod 8$

$$ \begin{array}{|c||c|c|c|} \hline & r_0N & r_1N & r_2N & r_3N & s_0N & s_1N & s_2N & s_3N \\ \hline \hline r_0N & r_0N & r_1N & r_2N & r_3N & s_0N & s_1N & s_2N & s_3N\\ \hline r_1N & r_1N & r_2N & r_3N & r_4N & s_1N & s_2N & s_3N & s_4N\\ \hline r_2N & r_2N & r_3N & r_4N & r_5N & s_2N & s_3N & s_4N & s_5N\\ \hline r_3N & r_3N & r_4N & r_5N & r_6N & s_3N & s_4N & s_5N & s_6N\\ \hline s_0N & s_0N & s_7N & s_6N & s_5N & r_0N & r_7N & r_6N & r_5N\\ \hline s_1N & s_1N & s_0N & s_7N & s_6N & r_1N & r_0N & r_7N & r_6N\\ \hline s_2N & s_2N & s_1N & s_0N & s_7N & r_2N & r_1N & r_0N & r_7N\\ \hline s_3N & s_3N & s_2N & s_1N & s_0N & r_3N & r_2N & r_1N & r_0N\\ \hline \end{array} $$

Since the quotient group is of order $8$, it must be isomorphic to some other group of the same order. I also noticed that this quotient group has one element of order $1$ ($r_0$), four elements of order $2$ ($s_0, s_1, s_2, s_3$), one element of order $4$ ($r_2$), and two elements of order $8$ ($r_1, r_3$).

What group of order $8$ is this quotient group isomorphic to? I hope all my work is correct. Thank you for your time and help!!!

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  • $\begingroup$ Also, is it the case that [ $ D_8 : \langle r_4 \rangle \ $ ] = $8$ ? Which is why there are $8$ cosets? $\endgroup$ Commented Nov 8, 2015 at 6:21

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Wow, that's a lot of work. You could have saved yourself some time by noting the following:

$r_4 = r_1^4$ (henceforth I will call $r_1$ simply "$r$", and $s_0$ by $s$-we can similarly replace $s_j$ by $r^js$), and:

$r^4s = sr^4$.

This, in turn, implies $r^4g = gr^4$ for any $g \in D_8$. Thus both elements of $\{r_0,r_4\} = \{e,r^4\}$ commute with every element of $D_8$.

In particular, if $y \in \{e,r^4\}$ we have:

$xyx^{-1} = yxx^{-1} = ye = y$, which makes it clear this group is normal.

It is not true the quotient has any elements of order $8$, in fact the coset $rN$ only has order $4$:

$(rN)^2 = r^2N \neq N$, since $r^2 \not\in \{e,r^4\}$.

$(rN)^3 = r^3N \neq N$, since $r^3 \not\in \{e,r^4\}$.

$(rN)^4 = r^4N = N$, since $r^4 \in \{e,r^4\}$.

I'll give you some hints as to which group you have of order $8$:

It has an element $rN$ of order $4$, and an element $sN$ of order $2$. These two elements do not commute. The subgroup $\langle sN\rangle$ is not normal.

The clincher is $(sN)(rN) = (rN)^{-1}(sN)$.

Your Cayley table isn't done right, you seem not to recognize that:

$r_4N = N\\r_1N = r_5N\\r_2N = r_6N\\r_3N = r_7N\\s_0N = s_4N\\s_1N = s_5N\\s_2N = s_6N\\s_3N = s_7N$.

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  • $\begingroup$ Thank you for taking the time to answer my question. Yeah I felt like my Cayley table had some elements in there that shouldn't be. I think I can take it from here. If not I will ask if and when I get stuck. $\endgroup$ Commented Nov 8, 2015 at 6:43
  • $\begingroup$ The Quaternion group? $\endgroup$ Commented Nov 8, 2015 at 6:46
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    $\begingroup$ In $Q_8$, every subgroup is normal (it also only has one element of order $2$). $\endgroup$ Commented Nov 8, 2015 at 6:59
  • $\begingroup$ I believe $D_8 / \langle r_4 \rangle$ is isomorphic to $D_4$ $\endgroup$ Commented Nov 10, 2015 at 7:26
  • $\begingroup$ That is correct. Can you prove it? $\endgroup$ Commented Nov 10, 2015 at 23:33

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