Abstract Nonsense

A Clever Proof of a Common Fact

Point of Post: In this post we give a new proof that if $G$ is a finite group $H$ is a subgroup $G$ whose index is the smallest prime dividing $|G|$ then that subgroup is normal.

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Motivation

It is a commonly used theorem in finite group theory that if $G$ is a finite group and $H\leqslant G$ such that $\left[G:H\right]=p$ is the smallest prime dividing $|G|$ then $H\unlhd G$. We have already seen a proof of this fact by considering the homomorphism $G\to S_p$ which is the induced map from $G$ acting on $G/H$ by left multiplication, and proving that $\ker(G\to S_p)=H$. We now give an even shorter (and the just mentioned proof is already short) proof of this fact using double cosets.

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September 16, 2011

The Character Table of S_3xZ_3

Point of post: In this post we use our recently developed theory on the irreducible representations (characters) of the products of groups to find the character table of a more formidable group, namely $S_3\times\mathbb{Z}_3$.

Motivation

Let’s put our newly developed theory to work.

April 11, 2011

Relation Between the Kernels of Characters and Normal Subgroups

Point of post: In this post we derive some fundamental relationships between the kernels of the characters of a finite group $G$ and the set of all normal subgroups of $G$. In particular, we prove that every normal subgroup of $G$ has the realization as the kernel of some character, and thus by previous theorem that every normal subgroup of $G$ has a realization as the intersection of certain $N^{(\alpha)}$‘s.

Motivation

We saw in our last post that there is a particularly fruitful way to produce normal subgroups ofa finite group $G$. Namely, one can look at the kernels of the characters of $G$, which as was shown amounts (in the sense that all the necessary information is contained in) to finding the kernels of the irreducible characters of $G$. What we shall now show is that in fact, finding the kernels of the irreducible characters of a group $G$ enables us to list off precisely what normal subgroups $G$ has. In particular, we will see that every normal subgroup of $G$ has a realization as the kernel of a certain character. We shall do this by constructing a certain character on the quotient group of $G$ for which, when thought of a character of $G$, will have kernel equal to the normal subgroup.

March 7, 2011

Review of Group Theory: Normal Subgroups and Quotient Groups (Pt. II)

Point of post: This post is a continuation of this one.

January 1, 2011

Review of Group Theory: Normal Subgroups and Quotient Groups (Pt. I)

Point of post: In this post we discuss several different formulations of normal subgroups and discuss how one can define a group structure on the set of cosets of such a normal subgroup.

Motivation

One sees the idea of a “quotient structure” pop up frequently in mathematics. There are quotient spaces in both topology and linear algebra for example. One mods out by the set-theoretic kernel of semi-metric to turn it into a metric etc. Of course, for groups the same holds. It turns out though that like algebra there are some pretty serious theorems to do with quotient groups (the isomorphism theorems) which we shall see in subsequent posts. The problem is that unlike both linear algebra (where all the underlying group structure of the vector spaces was assumed abelian) and topology the way we “want to” define a quotient group doesn’t always work; there is a necessary and sufficient condition on a subgroup such that the quotient group structure “makes sense”. This condition is normality which, in a nutshell, says that the subgroup is invariant under conjugation (or that it’s invariant under all inner automorphisms).

December 31, 2010

Topological Groups (Subgroups)

We will not extend our discussion of topological groups beyond the basic definitions and consequences relating to topological features to how “construction” features of the spaces (subspaces/subgroups, direct product/product space, etc.) react. For example, is a subgroup a topological group if given the subspace topology? That in particular will the the focus of the post, the answer being, yes. So, we begin by verifying this

Theorem: Let $G$ be a topological group and $H\leqslant G$. Then, $H$ is a topological group with the subspace topology.

Proof: It is clear that $H$ is a set with both a topological and group structure and thus it remains to show that $\vartheta:H\times H\to H:(h_1,h_2)\mapsto h_1h_2$ and $\varphi:H\to H:h\mapsto h^{-1}$ are continuous. It should be remarked that the only reason we know these two mappings map into their specified images is because $H$ is a subgroup. If it weren’t there’s no guarantee that $h_1,h_2\in H\implies h_1h_2=\vartheta(h_1,h_2)\in H$.

But, the continuity of these maps are apparent since $\varphi=\phi\mid_H$ and $\vartheta=\theta\mid_{H\times H}$ and since the restriction of a continuous map to a subspace is always continuous the conclusion follows. $\blacksquare$

Remark: As to not cause confusion I will always refer to the multiplication/inversion maps in the ambient topological groups as $\theta$ and $\phi$ and their subgroup analogues as the “theta variant” and “phi variant” $\vartheta$ and $\varphi$.

So, the natural question is when is a subgroup open or closed? Thus, we begin with some simple theorems relating to this:

Theorem: Let $H\leqslant G$. Then, $H$ is open if and only if it contains an interior point.

Proof: Suppose that it contains an interior point $h$. Then, there exists some $U$ such that $h\in U\subseteq H$. So, given any $h'\in H$ we have that $h'h^{-1}\in H$ and $h'h^{-1}h\in h'h^{-1}\subseteq h'h^{-1}H=H$ the last part gotten since multiplication of a subgroup by any of it’s elements merely permutes them, thus leaving the set essentially unchanged. Thus, it follows that $h'\in H^{\circ}$. But, since $h'$ was arbitrary it follows that $H$ is open.

Conversely, if $H$ is open then every point is an interior point, and since $H$ must contain $e$ it follows that $e$ is an interior point. The subtle point is that if $H$ weren’t a subgroup it very well could have been empty in which case it could still be open but not actually contain an interior point. $\blacksquare$

And in fact we have the following surprising theorem:

Theorem: Let $H\leqslant G$ be open, then it’s also closed.

Proof: If $H=G$ we’re done, so assume not and let $g\in G-H$. Then, $gH$ is a neighborhood of $g$ and $gH\cap H=\varnothing$ for if not then $gh_1=h_2$ for some $h_1,h_2\in H$ and so $g=h_2h_1^{-1}\in H$ which is a contradiction. It follows that $g\in \left(G-H\right)^{\circ}$. But, since it was arbitrary it follows that $G-H$ is open and thus $H$ is closed. $\blacksquare$

Corollary: Let $G$ be a connected topological group, then $G$ has no proper open subgroups.

Corollary: If you’ll remember earlier we proved that if $V$ is a symmetric neighborhood of $e$ then $\displaystyle \Omega=\bigcup_{n=1}^{\infty}V^n$ is a subgroup. But, since it’s evidently open (it’s the union of open sets) it follows from the previous corollary that if the ambient space $G$ is connected that it must be true that $\Omega=G$.

There is a partial converse too.

Theorem: Let $H\leqslant G$ be closed and $\left[G:H\right]<\infty$ then $H$ is open.

Proof: This follows easily since $G=H\cup g_1H\cup\cdots\cup g_n H$ for some $g_1,\cdots,g_n\in G$. But, all of these are closed and since they are disjoint (this is basic group theory) it follows that $H=G-\left(g_1H\cup\cdots g_nH\right)$ and the right hand side is $G$ minus the finite union of closed sets, thus open. The conclusion follows. $\blacksquare$.

Corollary: Let $H be closed. Then, if $G$ is connected it must be that $\left[G:H\right]=\infty$.

Remark: This is a convenient time to mention something really powerful about topological groups relating solely to group theory. Suppose that you wanted (I take a very simple example for convenience) to prove that $\mathbb{Z}\leqslant (\mathbb{R},+)$ has infinite index. All you’d need to do is pull a “in and out” technique. You note that you can topologize (going “in”) $(\mathbb{R},+)$ with the normal topology and this turns $\mathbb{R}$ into a connected topological group for which $\mathbb{Z}$ is a proper closed subgroup. It follows that $\left[\mathbb{R}:\mathbb{Z}\right]=\infty$. But, this is a purely group theoretic matter and so “going out” and forgetting the topology it still must be true that $\left[\mathbb{R}:\mathbb{Z}\right]=\infty$ since the topology has nothing to do with index. This will be a recurring technique one can implement. Namely, one wants to prove something about a group. So, you first see if you can topologize the group into a topological group where this quality must be true and then forget the topology.

We now give another example where the above remark can be used to great effect.

Theorem: Let $H\leqslant G$, then $\overline{H}\leqslant G$.

Proof: Clearly $e\in \overline{H}$. Now, let $h\in\overline{H}$ and let $N$ be any neighborhood of $h^{-1}$ . Then, $h\in N^{-1}$, but since $h\in\overline{H}$ there must exists some $h'\in N^{-1}\cap H$. So, $(h')^{-1}\in N$ but $(h')^{-1}\in H$. Thus, since $N$ was arbitrary it follows that $h^{-1}\in \overline{H}$.

Next, let $h_1,h_2\in\overline{H}$ and let $N$ be any neighborhood of $h_1h_2$. Remember that

$\theta^{-1}(N)=\bigcup_{j\in\mathcal{J}}\left(U_j\times V_j\right)$

where each $U_j,V_j$ is open. Thus,

$\displaystyle N=\theta(\theta^{-1}(N))=\theta\left(\bigcup_{j\in\mathcal{J}}\left(U_j\times V_j\right)\right)=\bigcup_{j\in\mathcal{J}}\theta\left(U_j\times V_j\right)=\bigcup_{j\in\mathcal{J}}U_jV_j$

Where each $U_jV_j$ being the product of open sets is open. But, since $\theta(h_1,h_2)=h_1h_2\in N$ it follows that $(h_1,h_2)\in U_{j_0}\times V_{j_0}$ for some $j_0\in\mathcal{J}$. But, this means that $U_{j_0}$ is a neighborhood of $h_1$ and thus it must contain some point $h'_1\in U_{j_0}\cap H$. Similarly, there must be some $h'_2\in V_{j_0}\cap H$. So,

$\displaystyle h'_1h'_2\in U_{j_0}V_{j_0}\subseteq\bigcup_{j\in\mathcal{J}}U_jV_j=N$

But, $h'_1h'_2\in H$. Thus, since $N$ was arbitrary it follows that $h_1h_2$ is an adherent point for $H$ and thus $H_1h_2\in\overline{H}$. $\blacksquare$

Theorem: Let $N\unlhd G$ then $\overline{N}\unlhd G$.

Proof: We have already shown that $\overline{N}\leqslant G$. Thus, it remains to show that $g\overline{N}g^{-1}\subseteq \overline{N}$ for every $g\in G$.

So, let $gng^{-1}\in g\overline{N}g^{-1}$ and let $U$ be any neighborhood of it. Then, $n\in g^{-1}Ug$, but this is a neighborhood of $n$ and so it must contain some point $n'\in N$. Thus, $gn'g^{-1}\in U$. But, since $N\unlhd G$ we know that $gn'g^{-1}\in N$. It follows $gng^{-1}$ is an adherent point and thus $gng^{-1}\in\overline{N}$. By previous comment the conclusion follows. $\blacksquare$

A nice theorem which is tangentially related is the following:

Theorem: Let $C\subseteq G$ be the component containing $C$. Then, it is closed and $C\unlhd G$.

Proof: Being closed is merely a consequence of the fact that a component in any topological space is closed.

Now, we prove that it’s a subgroup. To do this we need only prove that $C^{-1}\subseteq C$ and $C^2\subseteq C$.

Notice though that $e\in C^{-1}$ and since $\phi$ is a homeomorphism and $C^{-1}=\phi(C)$ it follows that $C^{-1}$ is connected. Thus, since $C\cap C^{-1}\ne\varnothing$ and both are connected it follows that $C\cup C^{-1}$ is connected. But, by the maximality of $C$ it follows that $C\cup C^{-1}=C\implies C^{-1}\subseteq C$ as desired.

Now, for closure under multiplication we note that $\displaystyle C^2=\bigcup_{c\in C}cC$. But, each $cC$ being the continuous image of $C$ (under $L_c$) is connected. But, since $c\in C\implies c^{-1}\in C\implies e\in cC^{-1}$ we see that $\displaystyle e\in\bigcap_{c\in C}cC$ and thus their intersection is non-empty. Thus, $\displaystyle \bigcup_{c\in C}cC=C^2$ is connected and contains $C$. It follows that $C^2=C$ as was desired.

Lastly, for normality we note (similarly as before that) $gCg^{-1}$ for any $g\in G$ is the continuous image of $C$ and thus connected. Also, $e\in gCg^{-1}$ so that $C\cup gCg^{-1}$ is a connected set containing $C$ and thus equal to $C$. It follows that $gCg^{-1}\subseteq C$. Since $g$ was arbitrary the conclusion follows. $\blacksquare$

Lastly, we prove  that a couple of specific kinds of subgroups of $G$ are closed. But, first a theorem.

Theorem:Let $\varphi,\psi:X\to G$ be continuous where $G$ is a topological group. Then, the map $\varphi\psi:X\to G:x\mapsto\varphi(x)\psi(x)$ is continuous.

Proof: One must merely note that

$\varphi\psi=\theta\circ\left(\varphi\oplus\psi\right):X\to G:x\overset{\varphi\oplus\psi}{\longmapsto}(\varphi(x),\psi(x))\overset{\theta}{\longmapsto}\varphi(x)\psi(x)$

And since both $\varphi\oplus\psi$ and $\theta$ are continuous the conclusion follows. $\blacksquare$.

So, with this we are able to prove our first theorem

Theorem: Recall from group theory that the centralizer $C_g$ is defined to be $C(g)=\left\{x\in G:gx=xg\right\}$. Then, $C(g)$ is closed in $G$ for every $g\in G$ if $G$ is Hausdorff.

Proof: From the above we know that $\phi C_g:G\to G:x\mapsto x^{-1}gxg^{-1}$ is continuous. And so

$C(g)=\left\{x\in G:xg=gx\right\}=\left\{x\in G:e=x^{-1}gxg^{-1}\right\}=\left(\phi C_g\right)^{-1}(\{e\})$

The conclusion follows. $\blacksquare$

From this we can prove the following theorem.

Theorem: Let $G$ be a topological group which is Hausdorff. Then, $\mathcal{Z}(G)$ (the center) is closed in $G$.

Proof: This follows since

$\displaystyle \mathcal{Z}(G)=\left\{x\in X:xg=gx\text{ }\forall g\in G\right\}=\bigcap_{g\in G}C(g)$

But, this is the intersection of closed sets and thus closed. $\blacksquare$

That’s it for now. Next time we’ll discuss “spaces of cosets”.

April 29, 2010