Normal subgroup test

Question

Hi there I have this problem: Is $ <p^6\epsilon^5> $ a normal subgroup of the Dihedral group $ D_4 = \{ I,p,p^2,p^3,\epsilon, p\epsilon, p^2\epsilon,p^3\epsilon \} $? Since I'm not that good at abstract algebra I only found that the group generated by $<p^6\epsilon^5>$ is the same as the group $<p^2\epsilon>$, which is flipping about the main diagonal, and the identity, right? Can you help me , please?

Answer 1

Hint: First verify that the subgroup $H = $ $<p^2\epsilon>$ is just $\{I, p^2\epsilon\}$ (looks like you already noticed that it has order two, although it isn't flipping around the main diagonal – if $\epsilon$ acts by flipping around one axis, $p^2\epsilon$ acts by flipping around the axis perpendicular to that). To determine whether $H$ is normal, we want to see if $ \forall x\in D_4$, $xHx^{-1} = H$. Check this with by hand, with $x=p$.

Answer 2

Let $S=\{p^6\epsilon^5,p^{12}\epsilon^{10},p^{18}\epsilon^{15},p^{24}\epsilon^{20},p^{30}\epsilon^{25},p^{36}\epsilon^{30},p^{42}\epsilon^{35},p^{48}\epsilon^{40}\}$

Note that the above list may contain repeats of the same elements.

Now, $p^4=I$ and $\epsilon^2=I$

$$S=\{p^2\epsilon,I,p^{2}\epsilon,I,p^{2}\epsilon,I,p^{2}\epsilon,I\}$$

$$S=\{I,p^2\epsilon\}$$

You should be able to verify that this is or isn't a normal subgroup.

Answer 3

Hint: Normality means that conjugation preserves the subgroup. But $$p^{-1}(p^2\epsilon)p = p\epsilon p = p(\epsilon p \epsilon)\epsilon = p(p^{-1})\epsilon = \epsilon.$$