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$1 \to ⟨ i ⟩ \to Q_{8} \to Q_{8} / ⟨ i ⟩ \to 1$ does not split, with nonabelian middle group

Statement refuted

Every short exact sequence of groups with cyclic kernel and cyclic quotient splits.

Let $Q_{8} = {\pm 1, \pm i, \pm j, \pm k}$ be the quaternion group (The quaternion group $Q_{8} = {\pm 1, \pm i, \pm j, \pm k}$ inside the nonzero quaternions) and $⟨ i ⟩ = {1, i, - 1, - i}$ . Then

$1 ⟶ ⟨ i ⟩ \overset{\subseteq}{⟶} Q_{8} \overset{π}{⟶} Q_{8} / ⟨ i ⟩ ⟶ 1$

is a short exact sequence whose kernel is cyclic of order $4$ and whose quotient is cyclic of order $2$ , and it is a counterexample: it has no section. Its middle group $Q_{8}$ is nonabelian, whereas the middle group of the cyclic witness $1 \to C_{p} \to C_{p^{2}} \to C_{p} \to 1$ has order $p^{2}$ and so is abelian; the two witnesses are therefore distinct.

Facts & Assumptions

Given: The quaternion group $Q_{8} \leq H^{\times}$ of The quaternion group $Q_{8} = {\pm 1, \pm i, \pm j, \pm k}$ inside the nonzero quaternions, with identity $1$ .

[L1]

For $Q_{8} = {1, - 1, i, - i, j, - j, k, - k} \subseteq H^{\times}$ : $Q_{8}$ is a subgroup of $H^{\times}$ with $∣ Q_{8} ∣ = 8$ ; $1$ is its only element of order $1$ , $- 1$ its only element of order $2$ , and each of $\pm i, \pm j, \pm k$ has order $4$ ; and $⟨ i ⟩ = {1, i, - 1, - i}$ is a subgroup of order $4$ containing $- 1$ ( $Q_{8}$ is a subgroup of $H^{\times}$ with eight elements, and $- 1$ is its only element of order $2$ ).

[L2]

A short exact sequence $1 \to N \overset{φ}{\to} G \overset{π}{\to} H \to 1$ consists of group homomorphisms with $φ$ injective, $π$ surjective and $im φ = \ker π$ ; a section is a homomorphism $s : H \to G$ with $π \circ s = {id}_{H}$ , and the extension splits when it has a section (Group extensions, sections, complements, and split extensions, The kernel and image of a group homomorphism).

[L3]

If $H \leq G$ and $[G : H] = 2$ , then $H ⊴ G$ (Every subgroup of index two is normal).

[L4]

For $N ⊴ G$ with $[G : N]$ finite, the quotient group $G / N$ is finite with $∣ G / N ∣ = [G : N]$ ; in particular $∣ G / N ∣ = ∣ G ∣ / ∣ N ∣$ when $G$ is finite (If $[G : N]$ is finite then $∣ G / N ∣ = [G : N]$ ; for finite $G$ this equals $∣ G ∣ / ∣ N ∣$ ).

[L5]

If $H \leq G$ with $G$ finite, then $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L6]

If $G$ is a finite group whose order is prime, then every $g \neq e$ has order $∣ G ∣$ and generates $G$ ; in particular $G$ is cyclic (A finite group of prime order is cyclic and every nonidentity element generates it).

[L7]

The order $ord (g)$ of an element $g$ of a group is the least natural $m \geq 1$ with $g^{m} = e$ when such an $m$ exists, and $\infty$ otherwise (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity, Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ); and a group homomorphism $f$ satisfies $f (e) = e^{'}$ (A group homomorphism automatically satisfies $f (e) = e^{'}$ and $f (g^{- 1}) = f (g)^{- 1}$ , and $f (g^{n}) = f (g)^{n}$ for every $n \in Z$ ; for monoid homomorphisms preservation of the identity must be assumed); the image of a group homomorphism is a subgroup of the codomain and its kernel is a normal subgroup of the domain (The image of a group homomorphism is a subgroup and its kernel is a normal subgroup).

[L8]

$i j = k$ and $j i = - k$ , and $k \neq - k$ ; so multiplication in $H$ is not commutative ( $H$ is a division ring that is not commutative, hence not a field: $q^{- 1} = \overset{ˉ}{q} / N (q)$ for $q \neq 0$ , while $i j = k$ and $j i = - k$ ).

[L11]

If $G$ is finite and $g \in G$ , then $g$ has finite order and $ord (g)$ divides $∣ G ∣$ (The order of every element of a finite group divides the order of the group).

[L10]

Every group of order $p^{2}$ , with $p$ prime, is abelian (Every group of order $p^{2}$ , for prime $p$ , is abelian).

[L9]

For $N ⊴ G$ the quotient group $G / N$ has the left cosets $g N$ as elements with product $(g N) (h N) = g h N$ ; the cosets form a group under it, whose identity is $N = e N$ and in which the inverse of $g N$ is $g^{- 1} N$ (The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ , For $N ⊴ G$ , the cosets form a group with identity $N$ and inverse $(g N)^{- 1} = g^{- 1} N$ ).

Counterexample

technique · contradiction

1.1

$⟨ i ⟩ ⊴ Q_{8}$ and $∣ Q_{8} / ⟨ i ⟩ ∣ = 2$ . By [L1] and [L5], $[Q_{8} : ⟨ i ⟩] = 8 / 4 = 2$ , so [L3] makes $⟨ i ⟩$ normal and [L4] gives $∣ Q_{8} / ⟨ i ⟩ ∣ = 2$ .

L1L3L4L5

1.2

The middle group $Q_{8}$ is nonabelian: $i j = k$ and $j i = - k$ with $k \neq - k$ by [L8], and $i, j, k, - k$ all lie in $Q_{8}$ by [L1]. The middle group of the cyclic witness $1 \to C_{p} \to C_{p^{2}} \to C_{p} \to 1$ has order $p^{2}$ and is therefore abelian by [L10], so the refutation below is not a restatement of that one.

L1L8L10

2.1

The projection $π : Q_{8} \to Q_{8} / ⟨ i ⟩$ , $π (g) = g ⟨ i ⟩$ , is a surjective homomorphism with $\ker π = ⟨ i ⟩$ . It is defined because $⟨ i ⟩$ is normal by step 1.1, it is a homomorphism because [L9] gives $(g ⟨ i ⟩) (h ⟨ i ⟩) = g h ⟨ i ⟩$ , it is surjective because every element of $Q_{8} / ⟨ i ⟩$ is a coset $g ⟨ i ⟩$ by [L9], and its kernel is ${g : g ⟨ i ⟩ = ⟨ i ⟩} = ⟨ i ⟩$ , the identity of the quotient being $⟨ i ⟩$ itself by [L9].

step 1.1L9algebra

2.2

Suppose, for contradiction, that a section $s : Q_{8} / ⟨ i ⟩ \to Q_{8}$ existed. Since $π \circ s = id$ , the map $s$ is injective, so its image $K = s (Q_{8} / ⟨ i ⟩)$ is a subgroup of $Q_{8}$ by [L7], of the same size as $Q_{8} / ⟨ i ⟩$ , hence $∣ K ∣ = 2$ by step 1.1. Writing $K = {1, x}$ with $x \neq 1$ , [L11] applied to the finite group $K$ gives that $ord (x)$ divides $∣ K ∣ = 2$ , so $ord (x)$ is $1$ or $2$ ; it is not $1$ , since by [L7] that would make $x = x^{1} = 1$ . Hence $ord (x) = 2$ .

step 1.1L2L7L11assume-contra

3.1

The sequence is short exact. Write $φ : ⟨ i ⟩ \to Q_{8}$ for the inclusion; it is an injective homomorphism with $im φ = ⟨ i ⟩$ , and step 2.1 gives $\ker π = ⟨ i ⟩$ , so $im φ = \ker π$ as [L2] requires. The kernel is $⟨ i ⟩$ , cyclic of order $4$ because it is generated by $i$ ; the quotient has order $2$ by step 1.1, and $2$ is prime, so [L6] makes it cyclic.

step 1.1step 2.1L1L2L6

3.2

By [L1] the only element of $Q_{8}$ of order $2$ is $- 1$ , so $x = - 1$ . But $- 1 \in ⟨ i ⟩ = \ker π$ by [L1] and step 2.1, hence $π (x)$ is the identity of $Q_{8} / ⟨ i ⟩$ .

step 2.1step 2.2L1L2

4.1

On the other hand $x = s (h)$ for the unique non-identity $h \in Q_{8} / ⟨ i ⟩$ , because $s$ is injective and $s$ sends the identity to the identity by [L7]; and then $π (x) = π (s (h)) = h \neq 1$ , contradicting step 3.2. Hence no section exists and the extension does not split. [step 2.2, step 3.2, L2, L7, discharge-contradiction] $■$

Depends on

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Sources

J. S. Milne, Group Theory, Example 3.9(c) (standard reference, not scraped)

1→⟨i⟩→Q8→Q8/⟨i⟩→11→⟨i⟩→Q8​→Q8​/⟨i⟩→1 does not split, with nonabelian middle group