AAR New algebraic objects from old: products and sums

Section 1.5 New algebraic objects from old: products and sums

With one of our goals to understand how a given algebraic object can be decomposed into simpler objects, it is necessary to understand how to build larger objects up from smaller ones. In one direction this leads to additional isomorphism theorems alluded to previously, but it also leads to things like direct and semi-direct products which are used to reveal the structure of many groups.

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As usual, we begin with the simplest of our algebraic objects under consideration, groups. Suppose that

H

and

K

are two subgroups of a group

G .

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Question 1.5.1.

A natural question is what is the smallest subgroup of $G$ which contains both $H$ and $K ?$

Answer.

Mathematics has an answer for us, but it may not be satisfying. The answer is that the smallest subgroup of $G$ containing both $H$ and $K$ is

\begin{equation*} \la H,K\ra := \bigcap_{J\supseteq H\cup K} J \end{equation*}

where the intersection is over all subgroups $J$ of $G$ which contain $H$ and $K\text{.}$

It is a very good answer in that it makes it clear such a group exists and is unique, but it gives us no idea how to construct it.

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Let's try again from a slightly different angle. If

J

is any subgroup of

G

which contains

H

and

K,

then since it is closed, it must contain all the products of the form

h k

and

k h

for

k \in K

and

h \in H .

We could ask a more naive question.

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Question 1.5.2.

Given subgroups $H$ and $K$ of a group $G,$ when is $H K := {h k ∣ h \in H, k \in K}$ a subgroup of $G,$ and perhaps when are $H K$ and $K H$ related?

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The answer to the question above is a well-known theorem.

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Theorem 1.5.3.

Let $H$ and $K$ be subgroups of a group $G .$ Assume that $K$ is a normal subgroup of $G,$ or more generally that $H$ contained in $N_{G} (K),$ the normalizer of $K .$ Then $H K = K H$ is a subgroup of $G,$ in particular the smallest subgroup of $G$ containing $H$ and $K .$

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An important fact to remember is the

H K = K H

does not mean that the elements of

H

and

K

commute, rather that for

h \in H, k \in K,

there exists

h^{'} \in H, k^{'} \in K

so that

h k = k^{'} h^{'} .

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Proof of Theorem 1.5.3.

The proof is just some simple manipulations using the concept of the normalizer. It should be clear that the identity is an element of $HK\text{.}$

Given $h,h^\prime \in H, k,k^\prime \in K\text{,}$ we need to know that $(hk)(h^\prime k^\prime) = h''k''$ for some $h''\in H$ and $k'' \in K.$ But

\begin{equation*} (hk)(h^\prime k^\prime) = h(h^\prime h^{\prime-1})k(h^\prime k^\prime) =(hh^\prime)(h^{\prime-1} k h^\prime)k^\prime = h''k'' \end{equation*}

where $h'' = hh^\prime$ and $k'' = (h^{\prime-1} k h^\prime)k^\prime\text{,}$ the last using $H\subseteq N_G(K).$

Given $h\in H, k \in K\text{,}$ we need to know that $(hk)^{-1} = h^\prime k^\prime$ for some $h'\in H$ and $k' \in K.$ But

\begin{equation*} (hk)^{-1} = k^{-1}h^{-1} = (h^{-1}h)k^{-1}h^{-1} = h^{-1}(h k^{-1}h^{-1}) = h'k'. \end{equation*}

Finally to show $HK = HK\text{,}$ we need to show the inclusions $HK \subseteq KH$ and $KH \subseteq HK.$ With predictable notation we see

\begin{align*} hk \amp = hk(h^{-1}h) = (hkh^{-1})h = k'h\\ kh \amp = (hh^{-1})kh = h (h^{-1}kh) = hk'. \end{align*}

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Example 1.5.4.

Let $G = S_{3},$ the symmetric group on three letters having order 6. Let $K = ⟨ (1 2 3) ⟩ = ⟨ (1 3 2) ⟩$ be the subgroup generated by either 3-cycle. We know this to be a normal subgroup the easiest reason being a consequence of Lagrange's theorem (reviewed in the next chapter). Let $H$ be any subgroup of order two (i.e., generated by any of the three transpositions. Then $H K$ is a subgroup of $G;$ indeed $G = H K,$ a fact we shall explore in the next chapter as well.

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Next we turn our attention to rings and their ideals.

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Definition 1.5.5.

Let $R$ be a ring with ideals $I$ and $J .$ We can define three new ideals from them: their sum and product and intersection. The sum, $I + J$ is the smallest ideal of $R$ which contains both $I$ and $J .$ The product, $I J,$ is the smallest ideal of $R$ which contains all the the elements of the form $i j$ with $i \in I$ and $j \in J .$ And of course the intersection, $I \cap J$ is the largest ideal contained in both.

While these properties define the ideals, the first two are not constructive definitions, but their characterization is not too hard to discern. One just has to ask how to make the generating sets closed under the operations of addition and multiplication by elements of the ring. One finds

\begin{aligned} I + J & = {i + j ∣ i \in I, j \in J} \\ I J & = {\sum_{k = 1}^{r} i_{k} j_{k} ∣ i_{k} \in I, j_{k} \in J} \\ I \cap J & = {k ∣ k \in I and k \in J} \end{aligned}

so the elements in $I J$ are finite sums of products of the form $i j .$

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Example 1.5.6.

Let $R = Z$ and let $I, J$ be ideals. We know that all the ideals of $Z$ are principal ideals, so let's look at three interesting cases.

$I = 12 Z$ and $J = 4 Z$
$I = 12 Z$ and $J = 15 Z$
$I = 12 Z$ and $J = 5 Z$

The resulting ideals $I + J, I J,$ and $I \cap J$ are:

$I + J = 4 Z;$ $I J = 48 Z;$ $I \cap J = 12 Z$
$I + J = 3 Z;$ $I J = 180 Z;$ $I \cap J = 60 Z$
$I + J = Z;$ $I J = 60 Z;$ $I \cap J = 60 Z$

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Checkpoint 1.5.7.

Let $R=\Z$ and let $I=m\Z$ and $J=n\Z$ be ideals.

Determine $I+J, IJ,$ and $I\cap J$ in terms of $m$ and $n\text{.}$ The expressions should be quite familiar to you.
Based upon the examples above, one might conjecture that
\begin{equation*} (I+J) \cdot (I\cap J) = IJ\text{.} \end{equation*}
Note that $(I+J) \cdot (I\cap J)$ is a product of ideals. Do you think it's always true for ideals of $\Z\text{?}$ It is not true in all rings.