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Given the Dihedral group $ D_4 $ (that is where $ D_4 = $ { $ id, R, R^{2}, R^{3}, F, RF, R^{2}F, R^{3}F $} ); Let $B =$ {$id, RF$}

I now wish to prove that $B$ is a subgroup of $D_4$:

  • Note that $B =$ {$id, RF$}, by definition of B we see it is 'non-empty'

  • $RF RF = id \in B$
    $ \therefore B$ is 'closed under operation'

  • $RF$ and $id$ are 'flips' and are their own inverses, thus:

    $\to$ $RF RF^{-1} = id \in B$

    $\to$ $id id^{-1} = id \in B$

    $ \therefore B$ is 'closed under inverses'

Now although i believe this holds true and accurately proves that $B$ is a subgroup of $D_4$ (correct me if I'm wrong though), i wish to show that $B$ is a subgroup using the definition of cyclic groups and how its properties can prove it's a subgroup. My definition of cyclic subgorup is:

  • "Cyclic Subgroup: if $a_1, a_2, .... ,a_n$ are any finite number of elements in $G$ (an arbitrary group), we define the subgroup generated '$a$' to by a cyclic subgroup (that's to say everything in this subgroup is a multiple or power of the generator) and is denoted $\langle a \rangle$"

any and all help is appreciated as i try to work through this !

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  • $\begingroup$ Use $\{{\rm id}\}$ for $\{{\rm id}\}$ and $\langle a\rangle$ for $\langle a\rangle$. $\endgroup$
    – Shaun
    Commented May 13, 2020 at 23:14
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    $\begingroup$ @Shaun will do! $\endgroup$ Commented May 13, 2020 at 23:15
  • $\begingroup$ This might come in handy. Also, are you familiar/comfortable with group presentations? I might be able to cook up an answer if so. $\endgroup$
    – Shaun
    Commented May 13, 2020 at 23:19
  • $\begingroup$ @Shaun I am vaguely familiar with them since i'm only at an introductory level of abstract algebra/group theory at the moment, but i understand the syntax and premise of them! $\endgroup$ Commented May 13, 2020 at 23:30

2 Answers 2

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Consider the subgroup $\langle RF\rangle$. You already said that you know $(RF)^2=id$. So now you have that the only elements in $\langle RF\rangle$ are $id$ and $RF$ so it is the same as your “set” $B$. So $B$ is a subgroup and a cyclic one at that.

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A presentation for $D_4$ in your notation would be

$$\langle F,R\mid R^4=F^2={\rm id}, FRF=R^{-1}\rangle.\tag{1}$$

Using Tietze transformations, we can introduce a generator $x$ with relation $x=RF$, i.e., $F=R^{-1}x$ to get $(1)$ equivalent to

$$\langle R,x\mid R^4=(R^{-1}x)^2={\rm id}, R^{-1}x^2=R^{-1}\rangle,\tag{2}$$

whose second relation gives $R=xR^{-1}x$ and whose third relation is equivalent to $x^2={\rm id}$, so that we get

$$\langle R,x\mid R^4={\rm id}, R^{-1}=xRx, x^2={\rm id}\rangle\tag{3}$$

from $(2)$. But note that $(3)$ is just like $(1)$; geometrically, $x$ is just another flip of the square other than $F$.

It follows that $\langle x\mid x^2\rangle$ (equivalent to your definition of a cyclic group) is a subgroup of $D_4$, which is what you're after.

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  • $\begingroup$ Does that make sense to you, @AJtheKiddd? I feel like I've skipped a few details. I work with presentations a lot, so I take short cuts. $\endgroup$
    – Shaun
    Commented May 13, 2020 at 23:55
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    $\begingroup$ wow thank you @Shaun! this took me a few reads through to grasp it all, but it definitely shows the inner workings of this idea very well and explicitly! i have an additional question upon reading this; to enumerate $D_{4/B}$ could it be written as $={[id],[R],[R^{2}],[R^{3}]}$ where each [ ] is a coset? (eg. $[R] = BR = {id, RF}R = {R, R^{2}F} $ ) $\endgroup$ Commented May 14, 2020 at 0:20
  • $\begingroup$ You're welcome, @AJtheKiddd! I'm happy that it helped. I'm not sure how to answer your follow-up question. I'll have to think about it tomorrow. (It's 01:28 here . . . now.) $\endgroup$
    – Shaun
    Commented May 14, 2020 at 0:28
  • $\begingroup$ no worries! i appreciate the help ! $\endgroup$ Commented May 14, 2020 at 0:32

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