# Using Definition of Cyclic Group to prove B is a Subgroup

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 !

• Use $\{{\rm id}\}$ for $\{{\rm id}\}$ and $\langle a\rangle$ for $\langle a\rangle$. May 13, 2020 at 23:14
• @Shaun will do! May 13, 2020 at 23:15
• This might come in handy. Also, are you familiar/comfortable with group presentations? I might be able to cook up an answer if so. May 13, 2020 at 23:19
• @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! May 13, 2020 at 23:30

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.

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.

• 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. May 13, 2020 at 23:55
• 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. $[R] = BR = {id, RF}R = {R, R^{2}F}$ ) May 14, 2020 at 0:20
• 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.) May 14, 2020 at 0:28
• no worries! i appreciate the help ! May 14, 2020 at 0:32