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The answer is "no". Let $G$ be a Tarski torsion-free monster of infinite exponent, $H$ be the trivial subgroup. Then $(H,G]$ consists of $G$ and all cyclic subgroups of $G$, it has ACC and infinite length. If $K$ is a cyclic subgroup and $a$ is not in the cyclic centralizer $C(K)$, then the coset $aK$ contains two non-commuting elements, so it generates $G$ but no coset of $H$ generates $G$.

The answer is "no". Let $G$ be a Tarski monster of infinite exponent, $H$ be the trivial subgroup. Then $(H,G]$ consists of $G$ and all cyclic subgroups of $G$, it has ACC and infinite length. If $K$ is a cyclic subgroup and $a$ is not in the cyclic centralizer $C(K)$, then the coset $aK$ contains two non-commuting elements, so it generates $G$ but no coset of $H$ generates $G$.

The answer is "no". Let $G$ be a Tarski torsion-free monster, $H$ be the trivial subgroup. Then $(H,G]$ consists of $G$ and all cyclic subgroups of $G$, it has ACC and infinite length. If $K$ is a cyclic subgroup and $a$ is not in the cyclic centralizer $C(K)$, then the coset $aK$ contains two non-commuting elements, so it generates $G$ but no coset of $H$ generates $G$.

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user6976
user6976

The answer is "no". Let $G$ be thea Tarski monster of infinite exponent, $H$ be the trivial subgroup. Then $(H,G]$ consists of $G$ and all cyclic subgroups of $G$, it has ACC and infinite length. If $K$ is a cyclic subgroup and $a$ is not in the cyclic centralizer $C(K)$, then the coset $aK$ contains two non-commuting elements, so it generates $G$ but no coset of $H$ generates $G$.

The answer is "no". Let $G$ be the Tarski monster of infinite exponent, $H$ be the trivial subgroup. Then $(H,G]$ consists of $G$ and all cyclic subgroups of $G$, it has ACC and infinite length. If $K$ is a cyclic subgroup and $a$ is not in the cyclic centralizer $C(K)$, then the coset $aK$ contains two non-commuting elements, so it generates $G$ but no coset of $H$ generates $G$.

The answer is "no". Let $G$ be a Tarski monster of infinite exponent, $H$ be the trivial subgroup. Then $(H,G]$ consists of $G$ and all cyclic subgroups of $G$, it has ACC and infinite length. If $K$ is a cyclic subgroup and $a$ is not in the cyclic centralizer $C(K)$, then the coset $aK$ contains two non-commuting elements, so it generates $G$ but no coset of $H$ generates $G$.

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user6976
user6976

The answer is "no". Let $G$ be the Tarski monster of infinite exponent, $H$ be the trivial subgroup. Then $(H,G]$ consists of $G$ and all cyclic subgroups of $G$, it has ACC and infinite length. If $K$ is a cyclic subgroup and $a$ is not in the cyclic centralizer $C(K)$, then anythe coset $aK$ contains two non-commuting elements, so it generates $G$ but no coset of $H$ generates $G$.

The answer is "no". Let $G$ be the Tarski monster of infinite exponent, $H$ be the trivial subgroup. Then $(H,G]$ consists of $G$ and all cyclic subgroups of $G$, it has ACC and infinite length. If $K$ is a cyclic subgroup and $a$ is not in the cyclic centralizer $C(K)$, then any coset $aK$ contains two non-commuting elements, so it generates $G$ but no coset of $H$ generates $G$.

The answer is "no". Let $G$ be the Tarski monster of infinite exponent, $H$ be the trivial subgroup. Then $(H,G]$ consists of $G$ and all cyclic subgroups of $G$, it has ACC and infinite length. If $K$ is a cyclic subgroup and $a$ is not in the cyclic centralizer $C(K)$, then the coset $aK$ contains two non-commuting elements, so it generates $G$ but no coset of $H$ generates $G$.

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user6976
user6976