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We know tha if $1\not=H\trianglelefteq G$ which consists of all the elements of $G$ which have finite order, then in $G/H$, no elements(except element neutral) has finite order.

My question is: If $1\not=H\trianglelefteq G$ ($H\not=G$) has exponent finite, then what can say on $G/H$?

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Unless I have misinterpreted your question, nothing can be said about $G/H$. Take for example $G= H \times K$, with $H$ finite (and hence of finite exponent) and $K$ arbitrary and non-trivial. $G/H$ is isomorphic to $K$ and this group can have any property you want.

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    @Lima: I think you might be onto something if $H$ was a *maximal* subgroup of a given exponent (as well as normal). Or, maybe rather, if it contained *all* elements of rank smaller than some number.2012-08-20