Quaternion Group has Normal Subgroup without Complement

Theorem

Let $Q$ denote the quaternion group.

There exists a normal subgroup of $Q$ which has no complement.

Proof

From Subgroups of Quaternion Group:

The subsets of $Q$ which form subgroups of $Q$ are:

\(\ds \)

\(\)

\(\ds Q\)

\(\ds \)

\(\)

\(\ds \set e\)

\(\ds \)

\(\)

\(\ds \set {e, a^2}\)

\(\ds \)

\(\)

\(\ds \set {e, a, a^2, a^3}\)

\(\ds \)

\(\)

\(\ds \set {e, b, a^2, a^2 b}\)

\(\ds \)

\(\)

\(\ds \set {e, a b, a^2, a^3 b}\)

From Quaternion Group is Hamiltonian we have that all of these subgroups of $Q$ are normal.

From Quaternion Group is Hamiltonian, all these subgroups are normal.

For two subgroups to be complementary, they need to have an intersection which is trivial.

However, apart from $\set e$ itself, all these subgroups contain $a^2$.

Hence none of these subgroups has a complement.

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $22$