- Show that, up to isomorphism, there is exactly one noncyclic group \(G\) of order \(4\).
- Show that \[ \operatorname{Aut}(G)\cong S_3. \]
Solution
Let \(e\) denote the identity of \(G\).
1. If \(G\) is noncyclic of order \(4\), every element of \(G\setminus\{e\}\)has order \(2\).
By Lagrange’s theorem, the order of a nonidentity element divides \(4\), so it is either \(2\) or \(4\). An element of order \(4\) would generate all of \(G\), contradicting the assumption that \(G\) is noncyclic. Hence every nonidentity element has order \(2\).
2. Up to isomorphism, the only noncyclic group of order \(4\) is\[ \ZZ/2\ZZ\times\ZZ/2\ZZ. \]
Choose distinct nonidentity elements \(a,b\in G\). By step 1, \(a^2=b^2=e\). The product \(ab\) is neither \(e\), \(a\), nor \(b\): the three possibilities would respectively imply \(a=b\), \(b=e\), or \(a=e\). Thus, if \(c\) denotes the third nonidentity element, then \(ab=c\).
Again by step 1, \(c^{-1}=c\). Therefore \[ ba =(ab)^{-1} =c^{-1} =c =ab, \] so \(G\) is abelian. The map \[ e\longmapsto(0,0),\qquad a\longmapsto(1,0),\qquad b\longmapsto(0,1),\qquad c\longmapsto(1,1) \] is then an isomorphism from \(G\) to \(\ZZ/2\ZZ\times\ZZ/2\ZZ\): it is bijective, and the relations \(a^2=b^2=e\), \(ab=ba=c\) imply \(c^2=e\), \(ac=b\), and \(bc=a\), so the displayed map preserves every product. This proves uniqueness up to isomorphism.
Set \[ \Omega\coloneqq G\setminus\{e\}. \] By step 2, \(\Omega\) has three elements.
3. Restriction to \(\Omega\) defines an injective homomorphism\[ \Phi:\Aut(G)\longrightarrow\operatorname{Sym}(\Omega)\cong S_3. \]
Every automorphism fixes \(e\) and therefore permutes the three elements of \(\Omega\), so restriction defines \(\Phi\). If an automorphism lies in the kernel of \(\Phi\), it fixes every element of \(\Omega\) and also fixes \(e\); hence it is the identity automorphism. Thus \(\Phi\) is injective.
4. Every permutation of \(\Omega\) extends to an automorphism of \(G\).
Let \(\sigma\in\operatorname{Sym}(\Omega)\) and extend it to \(G\) by setting \(\sigma(e)=e\). We verify that this extension preserves products.
If one factor is \(e\), preservation is immediate. If \(x=y\in\Omega\), then step 1 gives \[ \sigma(xy)=\sigma(e)=e=\sigma(x)^2. \] If \(x,y\in\Omega\) are distinct, the cancellation argument in step 2 shows that \(xy\) is the unique element of \(\Omega\setminus\{x,y\}\). The elements \(\sigma(x)\) and \(\sigma(y)\) are also distinct, and their product is the unique element of \(\Omega\setminus\{\sigma(x),\sigma(y)\}\), namely \(\sigma(xy)\). Thus \[ \sigma(xy)=\sigma(x)\sigma(y) \] in every case. Since the extension is bijective, it is an automorphism of \(G\).
5. One has\[ \boxed{\Aut(G)\cong S_3}. \]
Step 3 gives an injective homomorphism \(\Phi:\Aut(G)\to\operatorname{Sym}(\Omega)\), and step 4 shows that every permutation of \(\Omega\) is in its image. Hence \(\Phi\) is an isomorphism. Since \(\abs{\Omega}=3\), \(\operatorname{Sym}(\Omega)\cong S_3\).
6. Q.E.D.
Step 2 proves part (1), and step 5 proves part (2).