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If $G$ is a finite group and $H$ be its proper subgroup then $G =\bigcup gHg^{-1}$, where union varies over all $g$ belongs to $G$.

I thought like this :

It can be proved that if $G$ is a group then $gHg^{-1}$ is a subgroup of $G$. Hence $gHg^{-1}$ will give all possible subgroups of $G$. Hence, their union will give the whole group. Is my idea correct? If so, then, how can I prove it?... Thanks in advance.

Bach
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Math-Learner
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  • When you say “its proper subgroup” you imply that there is one and only one such proper subgroup; if you do not mean to imply that, then you shoudl say “a proper subbgroup”. Second, when you say “then”, do you mean to imply that $G$ will necessarily be the union of such conjugates? Because if that is the case, then you are wrong. – Arturo Magidin Sep 22 '19 at 18:48
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    What if $H$ is a normal subgroup?... – Trevor Norton Sep 22 '19 at 18:50
  • “Therefore, $gHg^{-1}$ will give all possible subgroups”. What could make you believe such a thing? For starters, $gHg^{-1}$ will always have the same size as $H$, and usually, not every proper subgroup has the same size. – Arturo Magidin Sep 22 '19 at 18:52
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    Moral: before spending time trying to prove a proposition, look at a few simple examples to see if it is true! – almagest Sep 22 '19 at 18:56
  • Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun Sep 22 '19 at 19:19

1 Answers1

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It is not true that every finite group is the union of conjugates of a proper subgroup. For example, if $G$ is abelian, then $gHg^{-1}=H$ for all $g\in G$, and therefore the union will just be $H$, and never all of $G$ (since $H$ is given as a proper subgroup). For example, $G=\mathbb{Z}/4\mathbb{Z}$, $H=2\mathbb{Z}/4\mathbb{Z}$.

Even if $G$ is not abelian, the result may fail. For example, if $H$ is normal, then $gHg^{-1}=H$ for all $g\in G$, so the union is never all of $G$. For example, $G=S_3$, $H=\langle (1,2,3)\rangle$.

Even if $H$ is not normal, the result may fail. For example, take $G=S_4$, and $H=\langle (1,2,3)\rangle$. Then every conjugate of $H$ lies in $A_4$, so $\cup_{g\in G}gHg^{-1}\subseteq A_4\neq S_4=G$ (in fact, the union will be all of $A_4$).

In fact, as noted in comments, a finite group is never the union of conjugates of the same proper subgroup; though that is perhaps not immediately obvious. You can find a discussion of this and what happens when $G$ is infinite here.

Arturo Magidin
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    Moreover, one can show that, if $H$ is a proper subgroup of $G$ and $G$ is finite, the equality $G=\bigcup_{g\in G} gHg^{-1}$ never holds. – Cristofer Villani Sep 22 '19 at 18:53