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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.2012-08-28
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    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](http://www.google.co.uk/url?sa=t&rct=j&q=varieties+of+groups&source=web&cd=1&ved=0CCgQFjAA&url=http%3A%2F%2Fwww.ams.org%2Fbull%2F1967-73-05%2FS0002-9904-1967-11795-6%2FS0002-9904-1967-11795-6.pdf&ei=dvk9ULeJBeOq0QW74IHYBg&usg=AFQjCNGBRmPqBn1u7PXg7MxxD8WdbascqA) lecture by BH Neumann will have to suffice.2012-08-29
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    (So, for example, all abelian groups satisfy the law $[x, y]=1$ while all groups of exponent $e$ satisfy the law $x^e$.)2012-08-29

3 Answers 3

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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).2012-08-28
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    Note that $H^G=H[G,H]$2012-08-29
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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!2012-08-30