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We know that if $H\leq G$, then $H^{G}=\left\langle h^{g};h\in H\text{ and }g\in G\right\rangle \trianglelefteq G,$ is the normal closure of $H$ in $G.$

Usually, when we kill $T\trianglelefteq G$ in $G/T$ , we have some property in $G/T.$

Example: If $T$ contains all the commutators of $G,$ then $G/T$ is abelian.

My question is: what can we say about the group $G/H^{G}$? What important properties does it have?

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All elements of H are trivial. –  i. m. soloveichik Aug 28 '12 at 14:24
    
You might like to look up varieties of groups. These are classes of groups which satisfy certain "laws" (and are not group varieties, as in algebraic groups). The place to start is a book by Hanna Neumann, but I can't link to that so this lecture by BH Neumann will have to suffice. –  user1729 Aug 29 '12 at 11:20
    
(So, for example, all abelian groups satisfy the law $[x, y]=1$ while all groups of exponent $e$ satisfy the law $x^e$.) –  user1729 Aug 29 '12 at 11:22
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3 Answers

up vote 1 down vote accepted

Let's generalize from your abelianization example. Commutators are elements of the form $aba^{-1}b^{-1}$, and a group being abelian means any element of the form $aba^{-1}b^{-1}$ is trivial. But there's nothing special about that form. You can write any form you like, such as $abc^2ba^{-1}$. If $H$ contains all elements of that form, then all such elements will be trivial in the quotient group.

This is closely related to the concept of a group presentation.

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I think you are describing “verbal subgroups” and “laws of groups” rather than group presentations. The difference is whether the a,b,c are variables that can be any group element (verbal and laws) or whether they are fixed generators of the group (normal subgroups and presentation). –  Jack Schmidt Aug 28 '12 at 21:22
    
Note that $H^G=H[G,H]$ –  Nicky Hekster Aug 29 '12 at 12:15
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There can be no special properties, since every normal subgroup $N$ arises in this way as $N^G$.

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One fact is the following. If $G$ is finite then $G$ must have a nilpotent subgroup $H$ with $H^G=G$. This can be seen by induction on $|G|$. If $G$ has a maximal subgroup $H$ that is not normal in $G$, then the result follows by induction applied to $H$. Otherwise all maximal subgroups are normal, which implies that $G$ itself is nilpotent.

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Lima: thank you very much! –  User2040 Aug 30 '12 at 21:48
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