What is $D_8/D_8$ isomorphic to? Why $\,1\,?\,$

By definition, I think it should be $D_8$, since $\,D_8/D_8=\{gD_8|g\in D_8\}=D_8$.

$(D_8/D_8:\;$ the quotient group of $D_8$ by itself.)

  • 6
    $\begingroup$ $\{D_8\}$ is a set with only one element, and $D_8$ is different from $\{D_8\}$. $\endgroup$
    – anon
    Mar 21, 2013 at 0:16
  • 7
    $\begingroup$ I find it strange that none of the answers have mentioned that quotient groups effectively shrink/collapse/identify elements of groups. All of the fancy terminology (Coset, normal subgroup, quotient, etc...) hides that very simple and important fact. $\endgroup$
    – Tyler
    Mar 21, 2013 at 0:19
  • 7
    $\begingroup$ @TylerBailey: You can add your own answer if you find the existing ones lacking, you know. $\endgroup$ Mar 21, 2013 at 0:25
  • $\begingroup$ I tried my best, let me know if it is incomprehensible or overly wordy! :p $\endgroup$
    – Tyler
    Mar 21, 2013 at 0:55
  • $\begingroup$ I'm voting to close this question as off-topic because it is very too old and very elementary $\endgroup$ Feb 6, 2020 at 10:59

5 Answers 5


The process of quotienting out a normal subgroup is just a fancy way of shrinking down the group we are working with. I'll try to give a non-technical answer because you seem bogged down by the details at the moment without an idea of what is actually happening:

Say you have a group $G$ and some (normal) subgroup $Q$. You don't want to deal with $G$ all by itself, it is too complicated. You decide to take ALL of the elements of $Q$ and consider them to be "the same." That is the quotient subgroup $G/Q$.

For your example, $D_8$: You decide to quotient by $D_8$, which means you are going to start considering any element I decide to give you to be all the same element. Then you're left with a group that only has $1$ element! It has to be the trivial group.

For another example, look at $\mathbb{Z} = \{...,-1, 0, 1, 2, ...\}$ and $2\mathbb{Z} = \{... ,-2, 0, 2, 4, ...\}$. We decide to quotient by $2\mathbb{Z}$. What do you think will happen? Well, we end up calling all of the EVEN elements one thing (we call evens $0$ since they have a nice property that $\mathrm{ E + E = E}$ and $\mathrm{ E + O = O}$ ), and all of the ODD elements another (call it $1$). All odd numbers are just an even number plus one anyway, right? So now we're down to a group $\mathbb{Z}/2\mathbb{Z}$ with two elements.

  • 2
    $\begingroup$ very nice explanation $\endgroup$
    – user27182
    Mar 21, 2013 at 14:56
  • $\begingroup$ I always appreciate the more wordy answers! $\endgroup$
    – user50229
    Apr 27, 2013 at 19:09
  • $\begingroup$ I think this is the way to do it, and as a bonus lets you see why normal subgroups are the ones you can quotient by. (They are kernels of "let's keep this the same" homomorphisms). I wrote a blog post about it! $\endgroup$
    – Eli Rose
    May 13, 2016 at 18:05

$G/G$ is the set of cosets of $G$ in itself under the operation of coset mulitplication. The quotient $G/G$ is the one element set $\{G\}$ with the operation of coset multiplication, and therefore isomorphic to the trivial group.


Hint: note that $$\left|\dfrac {D_8}{D_8}\right| = \left|D_8/D_8\right| = 1$$

What does that tell you with respect to what that one-element group $D_8/D_8$ must be?
And why, then, it must be isomorphic to the trivial group?

Try to better understand how "cosets" are defined and how quotient group is defined and how these relate, but also how they differ.


The error is that $\{ g D_8 | g \in D_8 \} = \{ D_8 \} \ne D_8$.

This follows from the fact that $1 D_8 = D_8$. All cosets are the same size as $D_8$ and subsets of $D_8$, so by the pidgeonhole principle, they must be equal.

But notice that $1 D_8 \in D_8 / D_8$. It is not equal to it.


$G/G$ is always equal to $1$. This is because the quotient group is the structure of the equivalency classes of cosets, not the cosets themselves. This is a group because we can inherit a group structure from $G$. So $D_8/D_8$ is a group of the cosets of $D_8$. There is only one of these.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .