Abstract Algebra – Nonnormal Subgroup of Q8 x Z4

abstract-algebradirect-productfinite-groupsgroup-theorynormal-subgroups

This problem from book Dummit and Foote "Abstract algebra":

Exhibit nonnormal subgroup of $Q_8 \times Z_4.$

Let $q \in Q_8$ and $z \in Z_4.$ Let $N$ be nonnormal subgroup of $Q_8 \times Z_4 $. Then there exist $g \in Q_8 \times Z_4$ and $n=(q_n,z_n) \in N$ such that $g^{-1}n g = (q^{-1},z^{-1}) (q_n,z_n) (q,z)=(q^{-1}q_nq,z^{-1}z_nz) \notin N.$
In the other hand every subgroup of $Q_8$ and every subgroup of $Q_8$ are normal.

I get tripped up on this problem!

Best Answer

Consider $N = \langle (i,1)\rangle$ which is cyclic of order $4$:

$(i,1)^2 = (-1,2)\\(i,1)^3 = (-i,3)\\(i,1)^4 = (1,0).$

Now $(j,0)(i,1)(j,0)^{-1} = (jij^{-1},0+1-0) = (ji(-j),1)$

$= ((-k)(-j),1) = (kj,1) = (-i,1) \not\in N$.