Quotient Groups - Dummit and Foote, Section 3.1, Exercise 17

[Math Amateur]

I am reading Dummit and Foote Section 3.1: Quotient Groups and Homomorphisms.

Exercise 17 in Section 3.1 (page 87) reads as follows:

-------------------------------------------------------------------------------------------------------------

Let G be the dihedral group od order 16.

$$G = < r,s \ | \ r^8 = s^2 = 1, rs = sr^{-1} >$$

and let $$\overline{G} = G/<r^4>$$ be the quotient of $$G$$ generated by $$r^4$$.

(a) Show that the order of $$\overline{G}$$ is 8

(b) Exhibit each element of $$\overline{G}$$ in the form $$\overline{s}^a \overline{r}^b$$------------------------------------------------------------------------------------------------------------------

I have a problem with part (b) in terms of how you express each element of $$\overline{G}$$ in the form requested - indeed, I am not quite sure what is meant by "in the form $$\overline{s}^a \overline{r}^b$$"My working of the basics of the problem was to put $$H = <r^4>$$ and generate the cosets of H as follows:

$$1H = H = \{ r^4, 1 \}$$

$$rH = \{ r^5, r \}$$

$$r^2H = \{ r^6, r^2 \}$$

$$r^3H = \{ r^7, r^3 \}$$

$$sH = \{ sr^4, s \}$$

$$srH = \{ sr^5, sr \}$$

$$sr^2H = \{ sr^6, sr^2 \}$$

$$sr^3H = \{ sr^7, sr^3 \}$$So the order of $$\overline{G}$$ is 8BUT - how do we express the above in the form $$\overline{s}^a \overline{r}^b$$ and what does the form mean anyway?

Would appreciate some help.

Peter

[Note: This has also been posted on MHF]

 

[Opalg]

I am reading Dummit and Foote Section 3.1: Quotient Groups and Homomorphisms.

Exercise 17 in Section 3.1 (page 87) reads as follows:

-------------------------------------------------------------------------------------------------------------

Let G be the dihedral group od order 16.

$$G = < r,s \ | \ r^8 = s^2 = 1, rs = sr^{-1} >$$

and let $$\overline{G} = G/<r^4>$$ be the quotient of $$G$$ generated by $$r^4$$.

(a) Show that the order of $$\overline{G}$$ is 8

(b) Exhibit each element of $$\overline{G}$$ in the form $$\overline{s}^a \overline{r}^b$$------------------------------------------------------------------------------------------------------------------

I have a problem with part (b) in terms of how you express each element of $$\overline{G}$$ in the form requested - indeed, I am not quite sure what is meant by "in the form $$\overline{s}^a \overline{r}^b$$"My working of the basics of the problem was to put $$H = <r^4>$$ and generate the cosets of H as follows:

$$1H = H = \{ r^4, 1 \}$$

$$rH = \{ r^5, r \}$$

$$r^2H = \{ r^6, r^2 \}$$

$$r^3H = \{ r^7, r^3 \}$$

$$sH = \{ sr^4, s \}$$

$$srH = \{ sr^5, sr \}$$

$$sr^2H = \{ sr^6, sr^2 \}$$

$$sr^3H = \{ sr^7, sr^3 \}$$So the order of $$\overline{G}$$ is 8BUT - how do we express the above in the form $$\overline{s}^a \overline{r}^b$$ and what does the form mean anyway?

Would appreciate some help.

Peter

[Note: This has also been posted on MHF]

You have essentially already done that when you listed the cosets. All you need to do is to notice that $\overline{s}^a \overline{r}^b$ is just an alternative notation for the coset $s^ar^bH$.

 

[Math Amateur]

You have essentially already done that when you listed the cosets. All you need to do is to notice that $\overline{s}^a \overline{r}^b$ is just an alternative notation for the coset $s^ar^bH$.

Thanks Opalg.

... just checking ...

It does not appear at first glance that $$r^3H = \{ r^7, r^3 \}$$ actually fits the form \(\displaystyle s^ar^bH \) unless of course we view it as \(\displaystyle s^0r^3H \).

Presumably that is the interpretation?

Peter

 

[Opalg]

Thanks Opalg.

... just checking ...

It does not appear at first glance that $$r^3H = \{ r^7, r^3 \}$$ actually fits the form \(\displaystyle s^ar^bH \) unless of course we view it as \(\displaystyle s^0r^3H \).

Presumably that is the interpretation?

Peter

Yes, you need to say that $r\in\{0,1,2,3\}$ and $s\in\{0,1\}$.

 

[Math Amateur]

Yes, you need to say that $r\in\{0,1,2,3\}$ and $s\in\{0,1\}$.

Opalg,

I do not follow why $r\in\{0,1,2,3\}$ and $s\in\{0,1\}$.

Can you explain?

Peter

 

[Opalg]

I do not follow why $r\in\{0,1,2,3\}$ and $s\in\{0,1\}$.

Can you explain?

Look at the list of cosets that you gave:

$$1H = \color{red}{s^0r^0H}$$
$$rH = \color{red}{s^0r^1H}$$
$$r^2H = \color{red}{s^0r^2H}$$
$$r^3H = \color{red}{s^0r^3H}$$
$$sH = \color{red}{s^1r^0H}$$
$$srH = \color{red}{s^1r^1H}$$
$$sr^2H = \color{red}{s^1r^2H}$$
$$sr^3H = \color{red}{s^1r^3H}$$

 

[Math Amateur]

Look at the list of cosets that you gave:

Hi Opalg,

From what you have emphasized regarding my calculations it appears that the cosets are of the form \(\displaystyle \overline{s}^a \overline{r}^b \)

where

\(\displaystyle a \in \{0,1 \} \)

and

\(\displaystyle b \in \{ 0,1,2, 3 \} \)

Can you please comment?

[Maybe I am being pedantic, but just want to make sure i understand what is being said ...]

Peter

 

[Deveno]

One can visualize $\langle r \rangle$ in the following "geometric" way:

Imagine a unit circle in the plane with eight distinguished "dots", starting at (1,0), and spaced around the unit circle so as to form a regular octagon. $r$ then can be represented as the rotation that takes each dot to the next one counter-clockwise.

Now what does "modding out $\langle r^4 \rangle$" correspond to?

You can imagine it like so: We deform the circle by "squeezing" the points (-1,0) and (1,0) together, to form TWO circles, like a figure 8, and then "fold the top one over" fusing the two circles. So now what used to be a 1/8 turn (on the original circle) is a 1/4 turn on the fused pair.

$s$ plays no part in this, all we have done with $G$ is to effectively replace the rotation generator $r$ with a new one $rH$ that now has an order of 4, instead of 8. In this quotient, the coset of $s$ is still of order 2 (because we have $s^2 = e \in H$ and $s \not\in H$, so that $sH \neq H$) and the relation:

$rs = sr^{-1}$

still holds in $G/H$:

$(rH)(sH) = rsH = sr^{-1}H = (sH)(r^{-1}H) = (sH)(rH)^{-1}$.

Given this, do you see how we can form a surjective homomorphism: $D_8 \to D_4$?

 

[Math Amateur]

One can visualize $\langle r \rangle$ in the following "geometric" way:

Imagine a unit circle in the plane with eight distinguished "dots", starting at (1,0), and spaced around the unit circle so as to form a regular octagon. $r$ then can be represented as the rotation that takes each dot to the next one counter-clockwise.

Now what does "modding out $\langle r^4 \rangle$" correspond to?

You can imagine it like so: We deform the circle by "squeezing" the points (-1,0) and (1,0) together, to form TWO circles, like a figure 8, and then "fold the top one over" fusing the two circles. So now what used to be a 1/8 turn (on the original circle) is a 1/4 turn on the fused pair.

$s$ plays no part in this, all we have done with $G$ is to effectively replace the rotation generator $r$ with a new one $rH$ that now has an order of 4, instead of 8. In this quotient, the coset of $s$ is still of order 2 (because we have $s^2 = e \in H$ and $s \not\in H$, so that $sH \neq H$) and the relation:

$rs = sr^{-1}$

still holds in $G/H$:

$(rH)(sH) = rsH = sr^{-1}H = (sH)(r^{-1}H) = (sH)(rH)^{-1}$.

Given this, do you see how we can form a surjective homomorphism: $D_8 \to D_4$?

Thanks for those helpful insights, Deveno.

I can see how the two circles come about from squeezing the points (-1,0) and (1,0) ... see my attached diagram and brief explanation ...

I am not sure what the above does to the octagon but inuitively it turns it into a sqare in each of the smaller circles, and hence the octagon becomes a square in the overlapped circles ... so I can imagine a transformation from \(\displaystyle D_8 \) to \(\displaystyle D_4 \) ... but exactly how to show this ... ?

You mention that if we replace the rotation generator r with rH where \(\displaystyle H = r^4 \), then rH has an order of 4. This seems right as

\(\displaystyle rH = \{ r^5, r \} \)

\(\displaystyle {rH}^2 = (rH)(rH) = r^2H = \{ r^6, r^2 \} \)

\(\displaystyle {rH}^3 = r^3H = \{ r^7, r^3 \} \)

\(\displaystyle {rH}^4 = r^4H = 1.H \)

Thus (rH) has an order of 4, compared with r which has an order of 8.
You also mention that the coset sH has an order of 2 ...

we have \(\displaystyle (sH)^2 = s^2H = 1H = H \)

So yes, sH has an order of 2 ...

(But what is the relevance of \(\displaystyle rs = sr^{-1} \)
Regarding a surjective homomorphism \(\displaystyle D_8 \to D_4 \) I cannot quite see how to do it ... can you help?

I can see that there is a surjective homomorphism: \(\displaystyle \pi \ : \ D_8 \to D_8/H \) where \(\displaystyle H = <r^4> \) given by

\(\displaystyle \pi (a) = aH \)

In this homomorphism for example:

\(\displaystyle \pi (r) = rH = \{r^5 , r \} \)

\(\displaystyle \pi (s) = sH = \{ sr^4 , s \} \)

and then

\(\displaystyle \pi (sr) = srH = \{ sr^5 , sr \} \)

So, check that \(\displaystyle \pi (s) \pi (r) = \pi (sr) \)

\(\displaystyle \pi (s) \pi (r) = \{ sr^4, s \}* \{ r^5, r \} \)

But ... how to multiply two sets ... ? ... Guess ... multiply all elements ...

Then get \(\displaystyle \pi (s) \pi (r) = \{ sr^9, sr^5, Sr^5, Sr \} = \{sr, sr^5 \} \)

which seems right ...

But how do we get a surjective homomorphism from \(\displaystyle D_8 \) to \(\displaystyle D_4 \)?

Would appreciate some help.

Peter

 

[Deveno]

The rule for multiplying two cosets is:

$(aH)(bH) = (ab)H$

For this to "make sense", we need to be confident that whenever:

$c \in aH$
$d \in bH$

that $cd \in (ab)H$.

If $H$ is normal, THIS ALWAYS MAKES SENSE.

Proof:

Suppose $H$ is normal, with $c \in aH, d \in bH$. Thus:

$c = ah$, for some element $h \in H$, and $d = bh'$ for some element $h' \in H$.

So... $cd = (ah)(bh')$.

But since $H$ is normal, $Hb = bH$, that is, $hb = bh''$ for some (possibly different) $h'' \in H$. Thus:

$cd = (ah)(bh') = a(hb)h' = a(bh'')h' = (ab)(h''h') \in (ab)H$

(since $h''h' \in H$ by the closure property of a subgroup).

*******

 

[Math Amateur]

The rule for multiplying two cosets is:

$(aH)(bH) = (ab)H$

For this to "make sense", we need to be confident that whenever:

$c \in aH$
$d \in bH$

that $cd \in (ab)H$.

If $H$ is normal, THIS ALWAYS MAKES SENSE.

Proof:

Suppose $H$ is normal, with $c \in aH, d \in bH$. Thus:

$c = ah$, for some element $h \in H$, and $d = bh'$ for some element $h' \in H$.

So... $cd = (ah)(bh')$.

But since $H$ is normal, $Hb = bH$, that is, $hb = bh''$ for some (possibly different) $h'' \in H$. Thus:

$cd = (ah)(bh') = a(hb)h' = a(bh'')h' = (ab)(h''h') \in (ab)H$

(since $h''h' \in H$ by the closure property of a subgroup).

*******

Thanks Deveno, yes, that is clear.

Can you help on the surjective homomorphism from \(\displaystyle D_8 \) to \(\displaystyle D_4 \)?

Peter

 

[Deveno]

Let's write:

$D_8 = \{1,r,r^2,r^3,r^4,r^5,r^6,r^7,s,sr,sr^2,sr^3,sr^4,sr^5,sr^6,sr^7\}$

Where $r^8 = s^2 = 1$.

The multiplication table is completely determined by the rule:

$rs = sr^{-1}$.

For example:

$(sr^2)(sr^3) = (sr)(rs)r^3 = (sr)(sr^{-1})r^3 = (srs)r^2$

$ = s(rs)r^2 = s(sr^{-1})r^2 = s^2r = r$

(It is faster when verifying the entire multiplication table of 256 entries (!) to first prove the rule:

$r^ms = sr^{-m}$)

Let us write:

$D_4 = \{e,a,a^2,a^3,b,ba,ba^2,ba^3\}$ where $a^4 = b^2 = e$,

subject to the rule:

$ab = ba^{-1}$.

Then, all we must do is verify that the following map gives a homomorphism:

$\phi(s^kr^m) = b^ka^m$

(There are four cases to check, I leave them to you).

Note that this mapping is NOT one-to-one, for example:

$\phi(r^2) = a^2 = ea^2 = (a^4)(a^2) = a^6 = \phi(r^6)$.

What is $\text{ker}(\phi)$?

If $\phi(s^kr^m) = b^ka^m = e$, we must have $k = 0$, that is:

$s^kr^m \in \langle r \rangle$.

Direct computation shows that the only $m$ for which this holds is:

$m = 0,4$, so $\text{ker}(\phi) = \{1,r^4\}$.

 

[Math Amateur]

Let's write:

$D_8 = \{1,r,r^2,r^3,r^4,r^5,r^6,r^7,s,sr,sr^2,sr^3,sr^4,sr^5,sr^6,sr^7\}$

Where $r^8 = s^2 = 1$.

The multiplication table is completely determined by the rule:

$rs = sr^{-1}$.

For example:

$(sr^2)(sr^3) = (sr)(rs)r^3 = (sr)(sr^{-1})r^3 = (srs)r^2$

$ = s(rs)r^2 = s(sr^{-1})r^2 = s^2r = r$

(It is faster when verifying the entire multiplication table of 256 entries (!) to first prove the rule:

$r^ms = sr^{-m}$)

Let us write:

$D_4 = \{e,a,a^2,a^3,b,ba,ba^2,ba^3\}$ where $a^4 = b^2 = e$,

subject to the rule:

$ab = ba^{-1}$.

Then, all we must do is verify that the following map gives a homomorphism:

$\phi(s^kr^m) = b^ka^m$

(There are four cases to check, I leave them to you).

Note that this mapping is NOT one-to-one, for example:

$\phi(r^2) = a^2 = ea^2 = (a^4)(a^2) = a^6 = \phi(r^6)$.

What is $\text{ker}(\phi)$?

If $\phi(s^kr^m) = b^ka^m = e$, we must have $k = 0$, that is:

$s^kr^m \in \langle r \rangle$.

Direct computation shows that the only $m$ for which this hold is:

$m = 0,4$, so $\text{ker}(\phi) = \{1,r^4\}$.

Most helpful indeed!

Working through the post carefully now!

Peter