There are five nonisomorphic groups of order 8. For each of those groups \(G\), find the smallest positive integer n such that there is an injective homomorphism \(\varphi: G\rightarrow S_n\).
Solution
1. For the cyclic group \(C_8\),\[ \mu(C_8)=8. \]
An \(8\)-cycle gives an embedding \[ C_8\hookrightarrow S_8, \] so \(\mu(C_8)\le8\).
Conversely, if \(C_8\) embeds in \(S_n\), then \(S_n\) contains an element of order \(8\). The order of a permutation is the least common multiple of its cycle lengths. For this least common multiple to be divisible by \(8\), at least one cycle length must be divisible by \(8\). Hence \(n\ge8\).
2. For \(C_4\times C_2\),\[ \mu(C_4\times C_2)=6. \]
The commuting permutations \[ a=(1\,2\,3\,4), \qquad b=(5\,6) \] generate a subgroup of \(S_6\) isomorphic to \(C_4\times C_2\), so the minimal degree is at most \(6\).
Suppose instead that \(C_4\times C_2\) embedded in \(S_n\) for some \(n\le5\). Its element of order \(4\) would map to a permutation of order \(4\). Such a permutation on at most five points must contain a \(4\)-cycle and therefore is a \(4\)-cycle together with at most one fixed point. The centralizer of a \(4\)-cycle in \(S_4\) or \(S_5\) is precisely the cyclic group generated by that \(4\)-cycle, of order \(4\). Since the image of the abelian group \(C_4\times C_2\) must lie in the centralizer of its order-\(4\) element, its image could have order at most \(4\), contradicting injectivity. Thus \(n\ge6\).
3. For the elementary abelian group \(E=C_2^3\),\[ \mu(E)=6. \]
The three disjoint transpositions \[ (1\,2),\qquad(3\,4),\qquad(5\,6) \] generate a subgroup of \(S_6\) isomorphic to \(C_2^3\), so \(\mu(E)\le6\).
Now suppose \(E\) acts faithfully on a set \(X\) with \(|X|\le5\). Since \(E\) is a \(2\)-group, every orbit has size \(1\), \(2\), or \(4\). Because \(E\) is abelian, for a transitive orbit \(E/H\) the kernel of the action on that orbit is exactly \(H\).
If there is an orbit of size \(4\), its stabilizer has order \(2\). Since at most one point remains, every other orbit is fixed, whose action kernel is all of \(E\). Hence the kernel of the full action contains that order-\(2\) stabilizer and is nontrivial.
If there is no orbit of size \(4\), then there are at most two orbits of size \(2\). Their stabilizers \(H_1,H_2\) each have order \(4\). Their intersection satisfies \[ |H_1\cap H_2| =\frac{|H_1||H_2|}{|H_1H_2|} \ge \frac{4\cdot4}{8}=2. \] Fixed points do not shrink the kernel. Thus again the full action has nontrivial kernel. Therefore no faithful action has degree at most \(5\), so \(\mu(E)=6\).
4. For the dihedral group \(D_8\) of order \(8\),\[ \mu(D_8)=4. \]
The symmetry action of a square on its four vertices gives a faithful embedding \[ D_8\hookrightarrow S_4. \] Thus \(\mu(D_8)\le4\). An embedding into \(S_n\) with \(n\le3\) is impossible because an injective homomorphism would identify \(D_8\) with a subgroup of \(S_n\), forcing \[ 8=|D_8|\mid |S_n|\le6, \] which is impossible. Hence \(\mu(D_8)=4\).
5. For the quaternion group \(Q_8\),\[ \mu(Q_8)=8. \]
The regular action gives a faithful embedding \[ Q_8\hookrightarrow S_8, \] so \(\mu(Q_8)\le8\).
Suppose \(Q_8\) acted faithfully on a set of cardinality less than \(8\). Decompose the action into transitive orbits. Each orbit has the form \(Q_8/H\). Since its size is less than \(8\), the stabilizer \(H\) is nontrivial. Every subgroup of \(Q_8\) is normal, so the kernel of the action on \(Q_8/H\) is exactly \(H\). Moreover, every nontrivial subgroup of \(Q_8\) contains the central element \(-1\): the unique subgroup of order \(2\) is \(\{1,-1\}\), and every subgroup of order \(4\) also contains \(-1\).
Therefore \(-1\) lies in the kernel of the action on every orbit, hence in the kernel of the whole action. This contradicts faithfulness. Thus every faithful permutation action of \(Q_8\) has degree at least \(8\), and so \(\mu(Q_8)=8\).
6. In summary,\[ \boxed{ \mu(C_8)=8,\quad \mu(C_4\times C_2)=6,\quad \mu(C_2^3)=6,\quad \mu(D_8)=4,\quad \mu(Q_8)=8. } \]