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Section 2.2 Understanding quotients and further isomorphism theorems

Proposition 1.3.6 gives equivalent conditions for a subgroup \(H\) of a group \(G\) to be normal. We also know that the kernel of any group homomorphism is a normal subgroup; indeed being normal is equivalent to being the kernel of some homomorphism, though that fact is not in and of itself all that useful. On the other hand, both conditions are themselves useful as we shall see in examples.

Example 2.2.1.

Let \(F\) be a field, and \(G = GL_n(F)\text{,}\) the general linear group (of invertible \(n\times n\) matrices with entries in \(F\)). From the point of view of rings, if \(R = M_n(F)\) is the ring of \(n\times n\) matrices with entries in \(F,\) then \(G = R^\times\text{,}\) the unit group of the ring \(R.\) Let \(K = SL_n(F),\) the special linear group, the subset of invertible matrices whose determinant is one.

Give two proofs that \(K\) is a normal subgroup of \(G\text{,}\) one using Proposition 1.3.6, and the other characterizing \(K\) as the kernel of a group homomorphism.

Solution. 1

Directly, one can use determinants both to show that \(K\) is a subgroup, but also that it is normal. For the normality part, let \(g\in G\) and \(k \in K.\) To check that \(gkg^{-1} \in K\text{,}\) one needs only observe that

\begin{equation*} \det(gkg^{-1}) = \det(g) \det(k) \det(g)^{-1} = \det (k) = 1 \end{equation*}

to show \(gkg^{-1} \in K\text{.}\) We have used the fact that \(\deg(g^{-1}) = \det(g)^{-1},\) and of course that \(\det g, \det k\) are nonzero scalars in \(F\text{,}\) which commute.

Solution. 2

A second solution is to recognize \(K\) as the kernel of a homomorphism. One that comes to mind is

\begin{equation*} \varphi: GL_n(F) \to F^\times \end{equation*}

given by \(\varphi(g) = \det g.\) Then \(K\) is obviously the kernel.

Example 2.2.2.

With the notation as above, show that \(GL_n(F)/SL_n(F)\) is an abelian group.

Solution.

Since \(K=SL_n(F)\) is a normal subgroup of \(G=GL_n(F)\text{,}\) we know at least that the quotient is a group. One could show it is abelian directly by showing that \(gK g'K = g'K gK\) since

\begin{equation*} gK g'K = g'K gK \iff gg'K = g'gK \iff g^{-1}g'^{-1}gg' \in K \end{equation*}

via determinants, but that is a bit grungy, and does not really leave us with a sense of what \(G/K\) looks like.

We return to the homomorphism

\begin{equation*} \varphi: GL_n(F) \to F^\times \end{equation*}

given by \(\varphi(g) = \det g.\) We already know the kernel, and a moment's thought convinces you that \(\varphi\) is surjective for given \(\alpha \in F^\times\text{,}\) \(\varphi\) takes the diagonal matrix \(\diag(\alpha,1, \dots, 1)\) to \(\alpha.\) Thus via the first isomorphism theorem,

\begin{equation*} GL_n(F)/SL_n(F) \cong F^\times \end{equation*}

the multiplicative group of a field which is certainly an abelian group.

By Theorem 1.5.3, we have seen the condition \(H \subseteq N_G(K)\) proves that \(HK\) is a subgroup of \(G\text{,}\) and also easily shows that \(K \trianglelefteq HK.\) For the rest, we try to let our theorems do the work.

Consider the homomorphism

\begin{equation*} \varphi: H \to HK/K \end{equation*}

which is the composition of natural homomorphisms given by

\begin{equation*} H\to HK \text{ via } h \mapsto h\cdot 1 \in HK \end{equation*}

and the projection

\begin{equation*} HK \to HK/K \text{ via } hk \mapsto hkK\text{,} \end{equation*}

so \(\varphi(h)= hK\text{.}\)

Since the coset \(hkK = hK,\) we see the map \(\varphi\) is surjective. For the kernel, we see that \(\varphi(h) = hK = K\) iff \(h \in H\cap K,\) and the first isomorphism theorem gives the result.

Remark 2.2.4.

Note that when \(H,K\) are finite subgroups, this theorem is stronger than Proposition 2.1.3 which computed the cardinality of \(HK\text{.}\)

Exercise 2.2.1.

Let \(F\) be a field, \(G = GL_n(F)\text{,}\) \(K = SL_n(F)\text{,}\) and \(H=D_n(F)\) the subgroup of diagonal matrices in \(G.\) You have already observed that \(K\) is a normal subgroup of \(G.\)

(a)

Show that \(H=D_n(F)\) is not a normal subgroup of \(G.\)

Solution.

Recall that the process of diagonalizing a matrix \(A\) (if possible) is one of finding an invertible matrix \(P\) so that \(P^{-1}AP=D\) is diagonal, which means (generally) \(PDP^{-1} = A\) is not diagonal. For a specific example,

\begin{equation*} \ba{rr}2\amp 1\\1\amp 1\ea \ba{rr}1\amp 0\\0\amp 2\ea\ba{rr}2\amp 1\\1\amp 1\ea^{-1} = \ba{rr}0\amp 2\\-1\amp 3\ea. \end{equation*}
(b)

Show that \(G=HK.\)

Solution.

Let \(A \in G=GL_n(F)\text{,}\) and let \(\alpha = \det A \in F^\times.\) Then

\begin{equation*} \diag(\alpha^{-1},1,\dots,1) A \in SL_n(F)\text{,} \end{equation*}

so

\begin{equation*} G = GL_n(F) \subseteq HK\text{,} \end{equation*}

and of course \(HK \subseteq G\text{,}\) which finishes the argument.

(c)

Apply the second isomorphism theorem to \(H,K.\)

Solution.

Since we know that \(K \normal G\text{,}\) the theorem applies, so we have

\begin{equation*} GL_n(F)/SL_n(F) = HK/K \cong H/(H\cap K) \cong F^\times. \end{equation*}

Of course we could have proven this directly with the determinant map using domain \(H.\)

It is simply refreshing to let our theorems do all the work in the proof of this theorem with a seemingly complicated statement. With \(H,K\) normal in \(G\text{,}\) it is immediate that \(H\normal K\text{,}\) so all the quotients \(G/H\text{,}\) \(G/K\text{,}\) and \(K/H\) are groups. Now we let the theorems take over.

By the fundamental homomorphism theorem, the natural surjective map (projection) \(\pi: G \to G/K\) factors through the quotient \(G/H\text{,}\) so essentially for free we are handed a natural (well-defined) surjective homomorphism

\begin{equation*} \pi_*: G/H \to G/K \end{equation*}

defined by \(\pi_*(gH) = gK\text{.}\) We need only ask for its kernel and apply the first isomorphism theorem. But

\begin{equation*} \ker \pi_* = \{gH \in G/H \mid gK = K\} = \{gH\in G/H\mid g\in K\} = K/H. \end{equation*}
Example 2.2.6.

A simple is to take \(G = \Z\text{,}\) so all subgroups are normal, and for positive integers \(m,n\text{,}\) let \(K = m\Z\) and \(H = mn\Z.\) Thus

\begin{equation*} (\Z/mn\Z) / (m\Z/mn\Z) \cong \Z/m\Z. \end{equation*}

The correspondence is straightforward:

\begin{align*} H'\amp\le G' \mapsto \varphi^{-1}(H')=\{g\in G\mid \varphi(g) \in H'\} \le G,\\ H\amp \le G \mapsto \varphi(H). \end{align*}

The one important fact to prove in order to establish the correspondence is one-to-one is to note that for \(H \le G\text{,}\)

\begin{equation*} \varphi^{-1}(\varphi(H)) = HK, \end{equation*}

but since \(K \le H,\) we have \(HK = H.\)

Example 2.2.8.

What are the subgroups of \(\Z/24\Z\text{?}\)

Answer.

Consider the natural surjective homomorphism \(\varphi: \Z \to \Z/24\Z,\) and use the correspondence theorem. There is a one-to-one correspondence between the subgroups of \(\Z/24\Z\) and the subgroups of \(\Z\) which contain \(24\Z\) (the kernel). But every subgroup of \(\Z\) is of the form \(n\Z\) for some \(n\text{,}\) and \(n\Z\supseteq 24\Z\) if and only if \(n \mid 24\text{.}\) So the subgroups of \(\Z\) containing \(24\Z\) are

\begin{equation*} d\Z \text{ for } d= 1, 2, 3, 4, 6, 8, 12, 24\text{,} \end{equation*}

so for these same \(d\) the subgroups of \(\Z/24\Z\) are \(d\Z/24\Z.\)