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I'm not sure what makes an answer to a question with several problems of very variable difficulty a good answer :)

Anyway:

there's no "3-Shelah" group. That is, every infinite group admits a subset $W$ such that $W^3\neq G$ and $|W|=G$. (Actually one can arrange $W\cup W^2\cup W^3\neq G$.)

By Zorn, let $G$ be an infinite group. Let $A$, by Zorn, be a maximal subset such that $1\notin A\cup A^2\cup A^3$. Write $B=\langle A\rangle$ and $L$ the subgroup of $G$ generated by $\{g^6:g\in G\}$; clearly $L$ is normal in $G$.

For every $g\in G\smallsetminus A$, the maximality implies that $$1\in (A\cup\{g\})\cup(A\cup\{g\})^2\cup (A\cup\{g\})^3.$$ Since $1\notin A\cup A^2\cup A^3$, this means that $$1\in \{g\}\cup Ag\cup gA\cup\{g^2\}\cup A^2g\cup AgA\cup gA^2\cup g^2A\cup gAg\cup Ag^2\cup\{g^3\}.$$ Hence one of the following holds: $g=1$ or $g^2=1$ $g^3=1$ or $g\in A^{-1}$, or $g^2\in A^{-1}$ or $g\in (A^2)^{-1}$.

Hence, $g^6\in B$ for all $g\in G$; equivalently, $L\subset B$. In particular, Therefore, in $G/L$, every element satisfies $x^6=1$. Since groups of exponent 6 are solvable, it follows that either $G=L$, or $G$ has a normal subgroup of index 2 or 3. In the last two cases, we define this subgroup as $W$. Otherwise, $B=G$, that is, $A$ generates $G$. In particular, since $G$ is infinite, $|A|=|G|$, so we put $A=W$.

I'm not sure what makes an answer to a question with several problems of very variable difficulty a good answer :)

Anyway:

there's no "3-Shelah" group. That is, every infinite group admits a subset $W$ such that $W^3\neq G$ and $|W|=G$. (Actually one can arrange $W\cup W^2\cup W^3\neq G$.)

Let $G$ be an infinite group. Let $A$, by Zorn, be a maximal subset such that $1\notin A\cup A^2\cup A^3$. Denote by $\langle A\rangle$ the subgroup generated by $A$, and $G^{(6)}$ the subgroup of $G$ generated by $\{g^6:g\in G\}$; clearly $G^{(6)}$ is normal in $G$.

For every $g\in G\smallsetminus A$, the maximality implies that $$1\in (A\cup\{g\})\cup(A\cup\{g\})^2\cup (A\cup\{g\})^3.$$ Since $1\notin A\cup A^2\cup A^3$, this means that $$1\in \{g\}\cup Ag\cup gA\cup\{g^2\}\cup A^2g\cup AgA\cup gA^2\cup g^2A\cup gAg\cup Ag^2\cup\{g^3\}.$$ Hence one of the following holds: $g=1$ or $g^2=1$ $g^3=1$ or $g\in A^{-1}$, or $g^2\in A^{-1}$ or $g\in (A^2)^{-1}$.

Hence, $g^6\in \langle A\rangle$ for all $g\in G$; equivalently, $G^{(6)}\subset \langle A\rangle$. In $G/G^{(6)}$, every element satisfies $x^6=1$. Since groups of exponent 6 are solvable, it follows that either $G=G^{(6)}$, or $G$ has a normal subgroup of index 2 or 3. In the last two cases, we define this subgroup as $W$. Otherwise, $\langle A\rangle=G$, that is, $A$ generates $G$. In particular, since $G$ is infinite, $|A|=|G|$, so we put $A=W$.

I'm not sure what makes an answer to a question with several problems of very variable difficulty a good answer :)

Anyway:

there's no "3-Shelah" group. That is, every infinite group admits a subset $W$ such that $W^3\neq G$ and $|W|=G$.

By Zorn, let $G$ be an infinite group. Let $A$, by Zorn, be a maximal subset such that $1\notin A^3$. Write $B=\langle A\rangle$ and $L$ the subgroup generated by $\{g^6:g\in G\}$; clearly $L$ is normal.

For every $g\in G$, the maximality implies that $1\in (A\cup\{g\})^3$. Since $1\notin A$, this means that $$1\in A^2g\cup AgA\cup gA^2\cup g^2A\cup gAg\cup Ag^2\cup\{g^3\}.$$ Hence, $g^3=1$ or $g^2\in A^{-1}$ or $g\in (A^2)^{-1}$.

Hence, $g^6\in B$ for all $g\in G$; equivalently, $L\subset B$. In particular, Therefore, in $G/L$, every element satisfies $x^6=1$. Since groups of exponent 6 are solvable, it follows that either $G=L$, or $G$ has a normal subgroup of index 2 or 3. In the last two cases, we define this subgroup as $W$. Otherwise, $B=G$, that is, $A$ generates $G$. In particular, since $G$ is infinite, $|A|=|G|$, so we put $A=W$.

I'm not sure what makes an answer to a question with several problems of very variable difficulty a good answer :)

Anyway:

there's no "3-Shelah" group. That is, every infinite group admits a subset $W$ such that $W^3\neq G$ and $|W|=G$. (Actually one can arrange $W\cup W^2\cup W^3\neq G$.)

By Zorn, let $G$ be an infinite group. Let $A$, by Zorn, be a maximal subset such that $1\notin A\cup A^2\cup A^3$. Write $B=\langle A\rangle$ and $L$ the subgroup of $G$ generated by $\{g^6:g\in G\}$; clearly $L$ is normal in $G$.

For every $g\in G\smallsetminus A$, the maximality implies that $$1\in (A\cup\{g\})\cup(A\cup\{g\})^2\cup (A\cup\{g\})^3.$$ Since $1\notin A\cup A^2\cup A^3$, this means that $$1\in \{g\}\cup Ag\cup gA\cup\{g^2\}\cup A^2g\cup AgA\cup gA^2\cup g^2A\cup gAg\cup Ag^2\cup\{g^3\}.$$ Hence one of the following holds: $g=1$ or $g^2=1$ $g^3=1$ or $g\in A^{-1}$, or $g^2\in A^{-1}$ or $g\in (A^2)^{-1}$.

Hence, $g^6\in B$ for all $g\in G$; equivalently, $L\subset B$. In particular, Therefore, in $G/L$, every element satisfies $x^6=1$. Since groups of exponent 6 are solvable, it follows that either $G=L$, or $G$ has a normal subgroup of index 2 or 3. In the last two cases, we define this subgroup as $W$. Otherwise, $B=G$, that is, $A$ generates $G$. In particular, since $G$ is infinite, $|A|=|G|$, so we put $A=W$.

I'm not sure what makes an answer to a question with several problems of very variable difficulty a good answer :)

Anyway:

there's no "3-Shelah" group. That is, every infinite group admits a subset $W$ such that $W^3\neq G$ and $|W|=G$.

By Zorn, let $G$ be an infinite group. Let $A$, by Zorn, be a maximal subset such that $1\notin A^3$. Write $B=\langle A\rangle$.

For every $g\in G$, the maximality implies that $1\in (A\cup\{g\})^3$. Since $1\notin A$, this means that $$1\in A^2g\cup AgA\cup gA^2\cup g^2A\cup gAg\cup Ag^2\cup\{g^3\}.$$ Hence, $g^3=1$ or $g^2\in A^{-1}$ or $g\in (A^2)^{-1}$.

Therefore, in $G/B$, every element satisfies $x^6=1$. Since groups of exponent 6 are solvable, it follows that either $G=B$, or $G$ has a normal subgroup of index 2 or 3. In the last two cases, we define this subgroup as $W$. Otherwise, $B=G$, that is, $A$ generates $G$. In particular, since $G$ is infinite, $|A|=|G|$, so we put $A=W$.

I'm not sure what makes an answer to a question with several problems of very variable difficulty a good answer :)

Anyway:

there's no "3-Shelah" group. That is, every infinite group admits a subset $W$ such that $W^3\neq G$ and $|W|=G$.

By Zorn, let $G$ be an infinite group. Let $A$, by Zorn, be a maximal subset such that $1\notin A^3$. Write $B=\langle A\rangle$ and $L$ the subgroup generated by $\{g^6:g\in G\}$; clearly $L$ is normal.

For every $g\in G$, the maximality implies that $1\in (A\cup\{g\})^3$. Since $1\notin A$, this means that $$1\in A^2g\cup AgA\cup gA^2\cup g^2A\cup gAg\cup Ag^2\cup\{g^3\}.$$ Hence, $g^3=1$ or $g^2\in A^{-1}$ or $g\in (A^2)^{-1}$.

Hence, $g^6\in B$ for all $g\in G$; equivalently, $L\subset B$. In particular, Therefore, in $G/L$, every element satisfies $x^6=1$. Since groups of exponent 6 are solvable, it follows that either $G=L$, or $G$ has a normal subgroup of index 2 or 3. In the last two cases, we define this subgroup as $W$. Otherwise, $B=G$, that is, $A$ generates $G$. In particular, since $G$ is infinite, $|A|=|G|$, so we put $A=W$.