Question

ABSTRACT AGEBRA

( quotient group )

(8) Show that every subgroup of the quaternion group Q is a normal subgroup of Q, and construct the Cayley table of each quotient group. Use this to classify all homomorphic images of the group Q.

Answer 1

Recall the quaternion group , where 1 is the identity element and the element e commutes with all the other elements of Q.

Q=<e,i,j,k: 72 = js = k2 = ijk = e, e' =1>

Also recall for any group and a subgroup
HCG
,
G: H = 2 => H
is normal in .

As
Q = 8
, any proper subgroup
HCQ
has order which divides 8. So possible orders of H are
1.2.4.
Now if
H = 4
then note that
Q: H = 2
, hence
HAQ
. Also if
H = 1
gives us
H= {1}
and thus also
HAQ
. SO enough to prove for H with
H = 2
. Note that since
H = 2
this gives us H is cyclic, and generated by an element of Q of order 2. And note that only element in Q of of order 2 is e. Thus the only posibility of H is
H= {1, e
. And since e commutes with all elements of Q, for any
9εο
we have
geg-1 = egg-1 = e EH
. Hence
HAQ
.

For
H= {1}
,
O = HO
, hence the Cayley table of is same as the Cayley table of Q.

For
H= {1, e
, note that
Q/H = {i, j, k, 1}
, where
= 2 = 2 = 1
,
ji=ij=k
,
kj = jk = i
and
ki = ik = j
. That is isomorphic to
Z/2Z eZ/2Z
.

For any subgroup H of Q, of order 4, we have
Q/H = 2
, hence Q/H is cyclic generated by an element a of order 2, that is isomorphiv to $\mathbb Z/2\mathbb Z$

Recall any homomorphic image of Q is isomorphic to , where K is the kernel of the map (by first isomorphism theorem). Now since K is the kernel we have
KcQ
be a subgroup. And since the image is a group we must have
KaQ
. Thus By our previous discurssion we have is either isomorphic to Q or isomorphic to , where  
H= {1, e
, that is isomorphic to
Z/2Z eZ/2Z
or isomorphic to $\mathbb Z/2\mathbb Z$ , in the case when
K = 4
.

Feel free to comment if you have any doubts. Cheers!