*Algebra*

## Section 2.5. Homomorphisms

Exercise 2.5.1 Suppose \(G\) is cyclic. Let \(x\) be a generator of \(G\). Let \(y \in G'\). Then \(y = \varphi(x^i)\) for some \(i\) since \(\varphi\) is surjective, and this equals \(\varphi(x)^i\) since \(\varphi\) is a homomorphism. This holds for arbitrary \(y\), therefore \(\varphi(x)\) generates \(G'\).

Suppose \(G\) is abelian. Let \(x, y \in G'\). Since \(\varphi\) is surjective, we can find \(a, b \in G\) with \(\varphi(a) = x, \varphi(b) = y\), whence \(xy = \varphi(a)\varphi(b) = \varphi(ab) = \varphi(ba) = \varphi(b) \varphi(a) = yx\). This holds for arbitrary \(x, y\), so \(G'\) is abelian.

Exercise 2.5.2 \(K\) and \(H\) both contain the identity, so \(K \cap H\) contains the identity. For all \(x \in K \cap H\) we have \(x \in K\) and \(x \in H\), so \(x^{-1} \in K\) and \(x^{-1} \in H\), so \(x^{-1} \in K \cap H\). Finally, for all \(x, y \in K \cap H\), we have \(x, y \in K\) and \(x, y \in H\), so \(xy \in K\) and \(xy \in H\), so \(xy \in K \cap H\). Therefore \(K \cap H\) is a subgroup. In the case where \(K\) is a normal subgroup of \(G\), for all \(x \in K \cap H\) and \(g \in H\), we have that \(gxg^{-1} \in K\) by virtue of the normality of \(K\) and \(gxg^{-1} \in H\) since \(x, g \in H\), therefore \(gxg^{-1} \in K \cap H\); so \(K \cap H\) is a normal subgroup of \(H\) (but not necessarily of \(G\)).

Exercise 2.5.6 Consider the elementary matrix of the first kind with \(1\) in position \((r, c)\), which we will notate as \(E^{r,c}\). Let \(M \in GL_n(\mathbb{R})\). Then \(E^{r,c}M\) is the matrix obtained by adding row \(c\) of \(M\) to row \(r\), while \(ME^{r,c}\) is the matrix obtained by adding column \(r\) of \(M\) to column \(c\). The two results are the same if and only if each element of row \(r\) has the same value added to it by both multiplications and each element of column \(c\) has the same value added to it by both multiplications. For all \(i \neq c\), the value added to \(M_{r,i}\) by left-multiplication is \(M_{c,i}\) and it is unaffected by right-multiplication, therefore all such \(M_{c,i}\) must vanish in order for \(M\) to commute with \(E^{r,c}\). For all \(i \neq r\), the value added to \(M_{i,c}\) by right-multiplication is \(M_{i,r}\) and it is unaffected by left-multiplication, therefore all such \(M_{i,r}\) must likewise vanish. That is, all off-diagonal elements in row \(c\) must vanish and all off-diagonal elements in column \(r\) must vanish. The element \(M_{r,c}\) has \(M_{c,c}\) added to it by left-multiplication and \(M_{r,r}\) added by right-multiplication, so those two elements must be equal. Since \(r,c\) can be chosen arbitrarily, it follows that all off-diagonal elements of \(M\) vanish and all on-diagonal elements are equal to each other, that is, \(M\) is a scalar matrix. All nonzero scalar matrices are indeed central and invertible, so such matrices for the centre of \(GL_n(\mathbb{R})\). Note that this proof works for any field.