This one has kicked my *ss. I am trying to find a subgroup of $\displaystyle Q_8 \times Z_4$ that is not normal.

Now, all subgroups of $\displaystyle Q_8$ are normal and all subgroups of $\displaystyle Z_4$ are normal. Then I need to find a subgroup of $\displaystyle Q_8 \times Z_4$ that is not normal.

The only case that is nontrivial is the subgroup $\displaystyle < i > \times (1, b)$. ( (1, b) where $\displaystyle b^2 = 1$ is the only subgroup of $\displaystyle Z_4$.)

Notation: I am defining

$\displaystyle (a, c) \cdot < i > \times (1, b) = ( a < i > ) \times (c, cb)$ where a is an arbitrary element of $\displaystyle Q_8$ and c is an arbitrary element of $\displaystyle Z_4$.

and

$\displaystyle < i > \times (1, b) \cdot (a, c) = ( < i > a ) \times (c, bc)$

Okay, so given an arbitrary a and c we have

$\displaystyle (a, c) \cdot < i > \times (1, b) = ( a < i > ) \times ( c, cb) $ $\displaystyle = (a < i > ) \times (c, bc )$

because $\displaystyle Z_4$ is Abelian.

Since $\displaystyle < i >$ is normal in $\displaystyle Q_8$ the above is equal to

$\displaystyle (< i > a ) \times (c, bc) = < i >\times (1, b) \cdot (a, c) = ( a, c ) \cdot < i > \times ( 1, b )$.

So $\displaystyle < i > \times (1, b)$ is normal in $\displaystyle Q_8 \times Z_4$ for all a, c.

Where have I gone wrong?

Thanks!

-Dan