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I have the following problem and I do not understand what they mean.

Let $G$ be a finite group, and $H$ a strict subgroup of $G$. Show that $G$ cannot be equal tho the union $\bigcup_{g\in G} gHg^{-1}$. Hint: if $n=card(G/H)$ show that $H$ has at most $n$ distinct conjugates.

I somehow don't understand what they mean with $n$ distinct conjugates, because we only had an action by conjugation but I don't think that this has to do somthing with the exercise, does it?

So it would be very helpful if someone could explain me what I have to do exaclty.

kabenyuk
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user123234
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    A conjugate of $H$ is a subgroup of $G$ of the form $gHg^{-1}$ for some $g\in G$. – Captain Lama Nov 06 '21 at 16:18
  • A conjugate of $H$ is a subgroup in the form $gHg^{-1}$ for some $g\in G$. $H$ having at most $n$ distinct conjugates means that the cardinality of the set of the subgroups of $G$ which are conjugates of $H$ is smaller or equal to $n$. –  Nov 06 '21 at 16:19
  • But does this have something to do with the conjugation action? Because I don't see why exactly we need n to be the cardinality of $G\setminus H$ and not of only H or G – user123234 Nov 06 '21 at 16:20
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    I have corrected the formula $G\setminus H$ to $G/H$, since we are talking about cosets here. – kabenyuk Nov 06 '21 at 16:33

2 Answers2

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Let $g\in G$ and $h\in H$ and set $g' = gh$. Then $$g'H(g')^{-1} = (gh)H(gh)^{-1} = g(hHh^{-1})g^{-1} = gHg^{-1},$$ so the conjugate $g'H(g')^{-1}$ is the same for all $g'\in gH$. Therefore, there are at most $\#G/\#H$ different conjugates $gHg^{-1}$. Moreover, every such conjugate contains the neutral element $e$ since $e\in H$ and $g e g^{-1} = g g^{-1} = e$.

So $$\#\bigcup_{g\in G}gHg^{-1} = 1 + \#\bigcup_{g\in G} (gHg^{-1}\setminus\{1\}) \leq 1 + \#G/\#H \cdot (\#H - 1) = \#G - (\frac{\#G}{\#H} - 1) < \#G.$$ In the second step we used that the number of cosets $gH$ is $\frac{\#G}{\#H}$ and that $\#(gHg^{-1}) = \#H$. In the last step we have used that $\frac{\#G}{\#H} > 1$ since $H$ is a proper subgroup of $G$.

azimut
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  • But does such a coset in $G/H$ has the form $gH$ for $g\in G$? – user123234 Nov 06 '21 at 16:32
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    @Wave yes, that is just the definition of $G/H$. Its elements are the left cosets $gH$ with $g\in G$. – azimut Nov 06 '21 at 18:08
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    @Wave I've modified my answer a bit to avoid the symbol $G/H$. – azimut Nov 06 '21 at 18:12
  • sorry I was in hurry, I try to solve it now and will ask you again if there are some problems. Sure at the end I will accept it. – user123234 Nov 06 '21 at 18:46
  • sorry I have a question, in the first part we shown that there are at most n different conjugates. But why do we need the second part? – user123234 Nov 06 '21 at 19:07
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    @Wave Your exercise asks you to "Show that $G$ cannot be equal tho the union $\bigcup_{g\in G} gHg^{-1}$." That is done in the second part, by showing that the sizes do not match. – azimut Nov 06 '21 at 20:13
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Meaning of distinct conjugates of $H$ in $G$ is as follows: Take $x,y$ in $G$. If the the two subsets (which are actually subgroups) of $G$ given by $xHx^{-1}$ and $yHy^{-1}$ are different from each other they are distinct conjugates. So either $xHx^{-1}$ has an element not present in $ yHy^{-1}$, or vice versa.

The hint says that if $x$ and $y$ belong to the same coset of $H$, then the conjugate subgroups by $x$ and $y$ will be the same.

We know that any conjugate of $H$ being isomorphicc (by an inner automorphism of $G$) to $H$ has the same number of elements as $H$. Being subgroups they have the identity element of $G$. So the number of elements in the union of conjugates is at most k (m-1) whre $k$ is the index of $H$ in $G$ and $m$ is the order of $H$.

From the proof of Lagrange's theorem we know that order of $G$ is $km$ which is more than $k(m-1)$ and hence the union of conjugates can not be the whole of $G$.

ANother imperfect way of seeing this (which does not always work but still insntructive) is as follows.

ASSUME that $G$ has an element $z$ of order some $r$ and $H$ does not have such an element. In that case no conjugate of $H$ can have element of order $r$ and hence their union will not include the element $z$.

For example if two (or more) different prime numbers divide the order of $G$, and if $H$ is a Sylow subgroup of $G$ for one prime, then conjugates of $H$ cannot contain any non-identity element of the Sylow subgroups for other prime numbers.

P Vanchinathan
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