# every subgroup of the quaternion group is normal

Show that every subgroup of the quaternion group is normal and find the isomorphism type of the corresponding quotient ?

I know that $Q_8$ has a subgroup $\langle i\rangle=\{1,i,-1,-i\}$, $\langle j\rangle=\{1,j,-1,-j\}$, $\langle k\rangle=\{1,k,-1,-k\}$, $\langle -1\rangle=\{1,-1\}$. So basically, I have to prove that everyone of these subgroups are a normal subgroup the isomorphism type of the corresponding quotient. Would anyone has an idea how I can get started. I want keep in mind that we have not gone over Lagrange's theorem yet and has not proved that every subgroup of index $2$ is normal.

• It is very odd that you are constructing and identifying quotient groups without knowing Lagrange's theorem. – David Hill Mar 1 '16 at 22:45
• In any case, to show directly that a subgroup $H\leq G$ is normal, you need to check that $gxg^{-1}\in H$ for every $g\in G$ and $h\in H$. If $H$ is cyclic (e.g. $H=\langle i\rangle=\{1,i,-i,-1\}$) you can take $x$ to be the generator. Also, if the generators of $G$ (e.g. $i$ and $j$) normalize $H$, then $H$ is normal. – David Hill Mar 1 '16 at 22:47
• @DavidHill Interestingly I am in an Algebra 1 class where they wait until almost the end of the semester to introduce Lagrange's theorem, and are expected to prove a lot of things "the long way" before learning the easy way. – jeffery_the_wind Nov 21 at 14:34

It's easy to show that $\langle -1\rangle$ is normal, because the elements $1$ and $-1$ commute with each element.

Consider $\langle i\rangle$; you just need to prove $xix^{-1}\in\langle i\rangle$, for every $x\in Q_8$, because you know $x1x^{-1}=1\in \langle i\rangle$ and $x(-1)x^{-1}=-1\in\langle i\rangle$; also $x(-i)x^{-1}=-xix^{-1}$, so $x(-i)x^{-1}\in\langle i\rangle$ as soon as $xix^{-1}\in\langle i\rangle$.

The statement is obviously true for $x=1,-1,i,-i$. Try with $j$: $$jij^{-1}=ji(-j)=-j(ij)=-jk=-i$$ Similarly, $$kik^{-1}=ki(-k)=-(ki)k=-jk=-i$$

Apply the same for $\langle j\rangle$ and $\langle k\rangle$.

On the other hand, you don't need to know special theorems for deducing that a subgroup $N$ of index $2$ in a group $G$ is normal. Indeed, the left cosets are $N$ and $gN=G\setminus N$ (where $g\notin N$), whereas the right cosets are $N$ and $Ng=G\setminus N$. So, for any $x\in N$, $xN=N=Nx$, and for $x\notin N$, $xN=G\setminus N=Nx$.

For each proper subgroup $H$ and for each $g \in G$ show that $gHg^{-1}=H$.

For example if $H=<i>$ and $g=j$ then $gHg^{-1}=\{ j*1*-j,j*i*-j, j*-1*-j, j*-i*-j\}$ and then do the calculations to show that this is still $H$.

You can take some shortcuts to avoid checking all $g \in G$. If $g \in H$ then $gHg^{-1}=H$ by closure. If $gHg^{-1}=H$ then you can show that $g^{-1}Hg=H$.

For the order 4 subgroups you can prove isomorphism of the quotient by counting the elements in the group. For example $G/<i>$ only has two elements and is therefore isomorphic to $\mathbb Z _2$.

The order 2 subgroup will make an order 4 quotient group so it's either $\mathbb Z_4$ or $D_2$. Look for order 4 elements to show that they are indeed isomorphic to $Z_4$.