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Let $ \mathcal{M} $ be the set of all subsets of finite group $G$ which have $p^{\alpha} $elements. Thus $ \mathcal{M} $ has $ {p^{\alpha}m \choose p^{\alpha}} $ elements.

Given $M_1 ,M_2 \in \mathcal{M} $ ($M_1$ is a subset of G having $p^{\alpha} $ elements, and likewise so is $M_2$ ), define $M_1 \sim M_2$ if there exists an element $g \in G$ such that $M_1 = M _2 g$.

It is immediate to verify that this defines an equivalence relation on $\mathcal{M}$. Let $\{M_1, M_2, ... , M_n\} $be such an equivalence class in $\mathcal{M} $. By our very definition of equivalence in $\mathcal{M}$, if $g\in G$, for each $i = 1, ... , n$; $M_i g = M_j$ for some $j$, $1\leq J\leq n$. Let , $ H = \{ g \in G | M_1 g = M_1 \}$.

-In book "Topics in Algebra", 2nd edition, By I.N. Herstein, Page 93.

Questions:

  1. Is $H$ a Normalizer?

  2. How Can I prove, that $ n =o(G)/o(H)$ ?

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  • $\begingroup$ The normalizer of $M$ is $\{g\in G | g M g^{-1}=M\}$ $\endgroup$ – Marcel Jan 15 '16 at 16:37
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Let me amplify Jim's answer:

Let $A$ be any subset of a group $G$, with $g \in G$. If we define:

$Ag = \{ag: a \in A\}$, then I claim $|A| = |Ag|$. To prove this, I will show that the map:

$f:A \to Ag$ given by $f(a) = ag$ is a bijection.

1) $f$ is injective: suppose $f(a) = f(a')$. Then $ag = a'g \implies (ag)g^{-1} = (a'g)g^{-1} \implies a = a'$.

2) $f$ is surjective: let $x \in Ag (= f(A))$. Then $x = ag$ for some $a \in A$, so $x$ has the pre-image under $f$ of $a$.

This shows that right-multiplication by any element of $G$ induces a mapping:

$G \to S_k$, where $k$ is the number of subsets of $G$ of any given cardinality. Indeed, this mapping is a homomorphism:

If $f$ is "right-multiplication by $g$", and $f'$ is "right-multiplication by $g'$", then $gg'$ induces the permutation $f\circ f'$ (we are following Herstein's convention that $f \circ f'$ means first do $f$, then $f'$, not the usual way).

In other words, right-multiplication by an element of $G$ permutes the subsets of $G$ of any given cardinality. Now $H$, in this situation, is the elements of $G$ which fix a certain subset of that given cardinality, in this case, $M_1$. It is not hard to see this is indeed a subgroup of $G$, and corresponds to the permutations in $S_k$ (here $k = \binom{p^{\alpha}m}{P^{\alpha}}$) that fix $1$ that are also in the image of $G$.

Jim's claim is that $M_1x = M_1y \iff Hx = Hy$.

He does a good job of showing that if $y \in Hx$, that $M_1y = M_1x$. So let's concentrate on the converse (the forward implication). If $M_1x = M_1y$, then $M_1xy^{-1} = M_1$, so that $xy^{-1} \in H$, and thus $Hx = Hy$.

This shows that the (set) mapping: $G/H \to [M_1]_{\sim}$ is bijective, so we have that this equivalence class has the same cardinality of $G/H$ (here, this is just the set of right cosets, since $H$ may not be normal in $G$), which is $[G:H] = |G|/|H|$.


Herstein's book is a wonderful introduction to Abstract Algebra, but his omission of group actions hurts the exposition, here. Both the class equation (where you have the "normalizer" idea) and this theorem are "special cases" of the Orbit-Stabilizer Theorem, which might help you with your conceptual difficulties.

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  • $\begingroup$ Thanks for your detail answer. $\endgroup$ – Mike SQ Jan 17 '16 at 7:02
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For 1, $H$ is typically called a stabilizer in the language of group actions. It does not make sense to talk about $H$ as a normalizer since the objects involved are only sets and not subgroups (we say $H$ normalizes a subgroups $K\leq G$ if $hKh^{-1}=K$ for every $h\in H$).

For 2, note that for every $i=1,\ldots, n$, there is a $g\in G$ such that $M_1g=M_j$ (this says the action on this equivalence class is transitive). Now, given such a $g$, note that $M_1hg=M_i$ for every $h\in H$. Hence, there are at least $|H|$ elements $g\in G$ such that $M_1g=M_i$. To see that this is the exact count, note that if $M_1g=M_i=M_1g'$, then $$M_1=M_ig^{-1}=(M_1g')g^{-1}=M_1g'g^{-1}.$$ This shows that $g'g^{-1}\in H$ and so $g'=hg$ for some $h\in H$.

To summarize, we have shown that for each $i=1,\ldots n$, there is exactly $|H|$ elements of $G$ mapping $M_1$ to $M_i$. As every element of $G$ sends $M_1$ to some $M_i$, we deduce that $|G|=n|H|$ and we're done.

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  • $\begingroup$ It's still possible to define the normalizer of a set, but it's more useful to talk about the normalizer of a subgroup, as this often yields important order information. $\endgroup$ – David Wheeler Jan 16 '16 at 19:01
  • $\begingroup$ Hill , Thanks for the answer. $\endgroup$ – Mike SQ Jan 17 '16 at 7:01
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Ans to 1. probably you are trying to follow the advice given on book and assumed $H$ is normalizer . But you can prove 2 without that.

Ans to 2.

Consider the right coset $Hg$, of $H$; if $x, y \in G$ are in that coset, then $y=hx$.

But then,$M_1 y=M_1(hx)=(M_1 h)x=M_1 x$ since $h \in H $, so $M_1h=M_1$.

So, if two elements are in the same coset of $H$, then they are in the same set $M_1$.

Now, Consider each distinct right coset $Hg$. For, $k \in Hg; M_1 k=M_1$ when $k \in Hg =H$.

Otherwise $M_1 k=M_i$.

For each distinct coset, there is an one-to-one relation with each $ M_i$.

So, The number of coset of $H = o(G)/o(H)=$ Total Number of $ M_i = n $.

Thus, $ n = o(G)/o(H)$.

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