Past exam of the mathematics course of the University of Cambridge / 2015 / ia / Paper 3 / 6D (Groups)

A group homomorphism satisfies $\phi (ab)=\phi (a)\phi (b)$ for every $a,b\in G$. Its kernel of a group homomorphism is $K=\{g:\phi (g)=e_H\}$. The homomorphism equation gives $\phi (e_G)=e_H$ by cancellation, and $\phi (g^{-1})=\phi (g{)}^{-1}$. Therefore $K$ contains the identity and is closed under multiplication and inverses, so it is a subgroup.

For $k\in K$ and $x\in G$,Thus conjugation carries $K$ into itself, and applying the inverse conjugation gives equality: $K$ is a normal subgroup. If $K=\{e,\xi \}$, conjugation cannot send $\xi$ to the identity because it is injective. ConsequentlyThis proves directly the normal subgroup of order two is central principle for kernels, including the kernel facts needed below.

True. The determinant gives a group homomorphismFor an orthogonal matrix, $A^{T}A=I$ implies $(\det A{)}^2=1$, and a reflection has determinant $-1$, so the target has order two and the map is onto. Its kernel of a group homomorphism is exactlyThis is the defining special orthogonal group inside the orthogonal group.

True. The sign of a permutation is a group homomorphism $\mathrm{sgn}:S_3\to \{1,-1\}$, as follows, for example, from the multiplicative determinant of permutation matrices. Its kernel of a group homomorphism is the alternating groupA transposition maps to $-1$, so this homomorphism is onto.

True. Use the odd-dimensional orthogonal determinant splitting:Scalar matrices commute with every matrix, so $\Phi (AB)=\Phi (A)\Phi (B)$. Its output is orthogonal andEvery $B\in SO(3)$ satisfies $\Phi (B)=B$, proving surjectivity. The equation $\Phi (A)=I$ forces $A=(\det A)I$, whenceIn odd dimension $-I$ has determinant $-1$, which is the sign needed in this construction.

False. Any supposed kernel of a group homomorphism of order two would have a nonidentity central element by the introductory proof. Every element of order two in $S_3$ is a transposition, and none is central: conjugating $(ij)$ by a suitable three-cycle gives a different transposition. Thus the center of a group for $S_3$ is trivial, and such a kernel is impossible. This argument does not assume that an arbitrary order-two subgroup is normal.