Subgroups of \(Q_8 \times E_{2^n}\) (D&F 5.1.6)

Show that all subgroups of \(Q_8 \times E_{2^n}\) are normal.

(Warning: potentially ugly/confusing working!)

The proof is by induction on \(n\). We first note 2 key properties of \(Q_8\): all subgroups are normal, and cosets of subgroups are conjugation-invariant (this can be verified manually). Let \(G = Q_8 \times Z_2\), and suppose \(H \le G\). We shall consider the restriction of the projection \(\pi_{Z_2} : G \rightarrow Z_2\) to \(H\): its image is a subgroup of \(Z_2\) and its kernel \(K\) is the set \(\{(q,1) \mid q \in Q\}\), where \(Q \le Q_8\). Using the First Isomorphism Theorem, we know \(K\) has index 1 or 2. If the former, then \(H = K \cong Q\), and for any \((g_1,g_2) \in G\) and \((h_1,h_2) \in H\) we have \((h_1,h_2)^{(g_1,g_2)} = (g_1h_1g_1^{-1},h_2) \in H\) (since \(Q \trianglelefteq Q_8\)).

Suppose the latter is true instead. Then \(H\) is partitioned into 2 cosets \(K\) and \(aK\), where \(a = (a_1,a_2)\) is any element with \(a_2 \neq 1\). It is easy to check that \(K \triangleleft G\), so it suffices to prove that \(aK\) is conjugation-invariant. If \((g_1,g_2) \in G\) and \((a_1b,a_2) = a(b,1) \in aK\), then \((a_1b,a_2)^{(g_1,g_2)} = (g_1(a_1b)g_1^{-1},a_2) \in aK\) (since \(a_1Q\) is conjugation-invariant). Thus \(H\) is normal. Furthermore, cosets of \(H\) are conjugation-invariant: let \(h = (h_1,h_2) \in H\), \(b = (b_1,b_2) \in G\) and \(g = (g_1,g_2) \in G\). Then \((bh)^g = (g_1(b_1h_1)g_1^{-1}, b_2h_2)\). But \(b_1h_1\) is in either \(b_1Q\) and \((b_1a_1)Q\) which are conjugation-invariant. Thus \((bh)^g \in bH\).

In conclusion, we have established the two properties mentioned at the beginning, but for \(Q_8 \times Z_2\). We can repeat the same argument for \(n = 2\) by replacing \(Q_8\) with \(Q_8 \times Z_2\), and so on.