# Finding what group a quotient group is isomorphic to without using Cayley diagrams.

Say we have the group $$D_8$$ and we want to find the quotient group of all of its normal subgroups which are (as $$D_8$$ is nilpotent) the maximal subgroups $$\langle r \rangle$$ $$,\langle r^2,s \rangle$$ $$\langle r^2,rs \rangle$$, and also as we know their intersection is normal $$\langle r^2 \rangle$$. The rest are found to be not normal by simple checking .

For an example take $$D_8$$/$$\langle r \rangle=\{\{r,r^2,r^3,e\},\{s,sr,sr^2,sr^3\}\}$$. I've been trying to find a good answer to why this group is isomorphic to $$C_2$$ ( and in general how to show that a given quotient group is isomorphic to a certain group) , But everywher I've looked has used Cayley diagrams without fully explaining how they've decided the mappings they're using in the diagram. I'm not good at reading Cayley diagrams or making them due to lack of practice anyway ( it was never something that we were at thought at college and is something I'm only now thinking I should learn by myself) .

My question is without resorting to Cayley diagrams how can we decide this quotient group is isomorphic to $$C_2$$ (or in general how can we mathematically decide without the use of Cayley diagrams what a quotient group is isomorphic to ?).

• $G = <r , s | r^4=s^2=1, srs = r^{-1}>$ so $G / <r> = <r , s | r=s^2=1, ss = 1> \cong < s | s^2=1>$ – reuns Aug 9 '19 at 0:19
• Any group with $2$ elements is isomorphic to $C_2$ – J. W. Tanner Aug 9 '19 at 0:19
• @J.W.Tanner what about larger quotient groups though ? – excalibirr Aug 9 '19 at 0:28
• @reuns I'm afraid I don't really see how that follows would you mind expanding a little ? – excalibirr Aug 9 '19 at 0:29
• $G= <r,s | r^4=s^2=(rs)^2= 1>$ means $G=Free(r,s) / < r^4,s^2,(rs)^2>$ (quotient of the free group by the subgroup generated by $r^4,s^2,(rs)^2$). Then $G/<r> \cong Free(r,s) / < r^4,s^2,(rs)^2,r>$. – reuns Aug 9 '19 at 0:33

We can check each quotient group one at a time.

First, let $$H = \langle r \rangle$$. Note that $$|H| = 4$$, and so $$|D_8/H| = 2$$. As pointed out in the comments, any group with two elements must be isomorphic to $$C_2$$. I'll give a proof of this fact at the bottom of this answer for completeness.

Let $$H = \langle r^2, s\rangle$$ or $$H = \langle r^2, rs\rangle$$. Since $$|H| = 4$$, we know that $$|D_8/H| = 2$$, and so $$D_8/H\cong C_2$$ in either case.

Let $$H = \langle r^2\rangle$$, so we have $$D_8/H = \{H, rH, sH, rsH\}$$. Note that $$(rH)(rH) = r^2H = H$$, so $$rH$$ has order $$2$$. We also know that $$(sH)(sH) = s^2H = H$$, so $$sH$$ also has order two. Since $$rsH = (rH)(sH)$$, we see that $$D_8/H = \langle rH, sH~|~ (rH)^2 = (sH)^2 = 1\rangle \cong C_2\times C_2$$. If this isn't obvious to you, you should prove this isomorphism yourself.

Here's the proof that any group of order $$2$$ must be $$C_2$$. You should really try to prove this yourself, but I'll give the proof (hidden, of course!).

Let $$G$$ be a group of order two, where $$G = \{1, b\}$$. Since $$b$$ must have an inverse, and $$1$$ can't be this inverse, it must be that $$b = b^{-1}$$, or that $$b^2 = 1$$. Thus $$G = \langle b~|~ b^2 = 1\rangle$$, i.e. a cyclic group of order two. Thus $$G \cong C_2$$.

The quotient groups (homomorphic images) that are nontrivial have order $$4$$ (there's one) and $$2$$ (there are three).

There is only one group of order two. For instance, $$2$$ is prime, and there is only one group of order $$p$$ for any $$p$$ prime, by Lagrange. That is the cyclic one.

As for the quotient group of order $$4$$, all the groups of order $$4$$ are abelian (the smallest nonabelian group is $$S_3\cong D_3$$). Then we know by FTFAG, that we have either $$\Bbb Z_4$$ or $$\Bbb Z_2×\Bbb Z_2$$. But it's easy to check that there is no element of order $$4$$.

For larger groups this problem can get more involved. See this. The OEIS sequence corresponding to the number of groups of size $$n$$ is A000001.