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Not always. Here is an answer in the realm of Lie algebras, and below I'll make it an answer in the realm of $p$-groups (a counterexample of order $p^{10}$ and class $5$ (and exponent $p$) in which every proper subgroup has class $\le 3$).

Fix an arbitrary ground field. Consider the vector space of basis $(e_i)_{1\le i\le 10}$. Make it a Lie algebra with the brackets (using the shortcut $i.j|x$ to mean $[e_i,e_j]=x=-[e_j,e_i]$, and other brackets being meant to be zero):

\begin{gather*} 1.2|e_3,\qquad 1.3|e_4,\;2.3|e_5,\qquad 1.4|e_6,\;1.5|e_7,\;2.4|e_7,\;2.5|e_8 \\ 1.7|e_9,\;2.7|-2e_9,\;1.8|e_{10},\;2.7|-2e_{10}. \end{gather*}

Then this satisfies Jacobi, and also satisfies that $[x,[x,[x,[x,y]]]]=0$ for all $x$, $y$ in the Lie algebra; the latter making use of the choice of $-2$ coefficients.

This is also a (Carnot) graded Lie algebra with $e_1$, $e_2$ of degree $1$, $e_3$ of degree $2$, $e_4$, $e_5$ of degree $3$, $e_6$, $e_7$, $e_8$ of degree $4$, and $e_9$, $e_{10}$ of degree $5$.

One then sees that the natural action of $\mathrm{GL}_2$ in the degree 1 space extends to the whole Lie algebra.

(Well, this Lie algebra was precisely produced to satisfy this, namely starting with the 2-generator free 5-step-nilpotent Lie algebra, observing that the degree 5 component splits as $\mathrm{GL}_2$-module as sum of a 4-dimensional and a 2-dimensional module, and killing the 4-dimensional module. The resulting Lie algebra is the free Lie algebra in the variety of 5-step-nilpotent Lie algebras satisfying the additional law $[x,[x,[x,[x,y]]]]=0$.)

Hence, all codimension-1 subalgebras of the above Lie algebra are isomorphic. Hence it is enough to see that a single one, say the one with basis $(e_i)_{2\le i\le 10}$ is 3-step nilpotent. Namely, its derived subalgebra has basis $(e_5,e_7,e_8,e_9,e_{10})$ and the next step in the central series has basis $(e_8,e_{10})$ and is central therein.

Now consider this Lie algebra over $\mathbf{Z}/p\mathbf{Z}$ with $p>5$. Then Lazard proved that the Baker–Campbell–Hausdorff formula produces a group law for which the lower central series coincides with the Lie-algebraic one, and subgroups also coincide with subalgebras. Thus the resulting group works.

Not always. Here is an answer in the realm of Lie algebras, and below I'll make it an answer in the realm of $p$-groups (a counterexample of order $p^{10}$ and class $5$ (and exponent $p$) in which every proper subgroup has class $\le 3$).

Fix an arbitrary ground field. Consider the vector space of basis $(e_i)_{1\le i\le 10}$. Make it a Lie algebra with the brackets (using the shortcut $i.j|x$ to mean $[e_i,e_j]=x=-[e_j,e_i]$, and other brackets being meant to be zero):

\begin{gather*} 1.2|e_3,\qquad 1.3|e_4,\;2.3|e_5,\qquad 1.4|e_6,\;1.5|e_7,\;2.4|e_7,\;2.5|e_8 \\ 1.7|e_9,\;2.6|-2e_9,\;1.8|e_{10},\;2.7|-2e_{10}. \end{gather*}

Then this satisfies Jacobi, and also satisfies that $[x,[x,[x,[x,y]]]]=0$ for all $x$, $y$ in the Lie algebra; the latter making use of the choice of $-2$ coefficients.

This is also a (Carnot) graded Lie algebra with $e_1$, $e_2$ of degree $1$, $e_3$ of degree $2$, $e_4$, $e_5$ of degree $3$, $e_6$, $e_7$, $e_8$ of degree $4$, and $e_9$, $e_{10}$ of degree $5$.

One then sees that the natural action of $\mathrm{GL}_2$ in the degree 1 space extends to the whole Lie algebra.

(Well, this Lie algebra was precisely produced to satisfy this, namely starting with the 2-generator free 5-step-nilpotent Lie algebra, observing that the degree 5 component splits as $\mathrm{GL}_2$-module as sum of a 4-dimensional and a 2-dimensional module, and killing the 4-dimensional module. The resulting Lie algebra is the free Lie algebra in the variety of 5-step-nilpotent Lie algebras satisfying the additional law $[x,[x,[x,[x,y]]]]=0$.)

Hence, all codimension-1 subalgebras of the above Lie algebra are isomorphic. Hence it is enough to see that a single one, say the one with basis $(e_i)_{2\le i\le 10}$ is 3-step nilpotent. Namely, its derived subalgebra has basis $(e_5,e_7,e_8,e_9,e_{10})$ and the next step in the central series has basis $(e_8,e_{10})$ and is central therein.

Now consider this Lie algebra over $\mathbf{Z}/p\mathbf{Z}$ with $p>5$. Then Lazard proved that the Baker–Campbell–Hausdorff formula produces a group law for which the lower central series coincides with the Lie-algebraic one, and subgroups also coincide with subalgebras. Thus the resulting group works.

Not always. Here is an answer in the realm of Lie algebras, and below I'll make it an answer in the realm of $p$-groups (a counterexample of order $p^{10}$ and class $5$ (and exponent $p$) in which every proper subgroup has class $\le 3$).

Fix an arbitrary ground field. Consider the vector space of basis $(e_i)_{1\le i\le 10}$. Make it a Lie algebra with the brackets (using the shortcut $i.j|x$ to mean $[e_i,e_j]=x=-[e_j,e_i]$, and other brackets being meant to be zero):

\begin{gather*} 1.2|e_3,\qquad 1.3|e_4,\;2.3|e_5,\qquad 1.4|e_6,\;1.5|e_7,\;2.4|e_7,\;2.5|e_8 \\ 1.7|e_9,\;2.7|-2e_9,\;1.8|e_{10},\;2.7|-2e_{10}. \end{gather*}

Then this satisfies Jacobi, and also satisfies that $[x,[x,[x,[x,y]]]]=0$ for all $x$, $y$ in the Lie algebra; the latter making use of the choice of $-2$ coefficients.

This is also a (Carnot) graded Lie algebra with $e_1$, $e_2$ of degree $1$, $e_3$ of degree $2$, $e_4$, $e_5$ of degree $3$, $e_6$, $e_7$, $e_8$ of degree $4$, and $e_9$, $e_{10}$ of degree $5$.

One then sees that the natural action of $\mathrm{GL}_2$ in the degree 1 space extends to the whole Lie algebra.

(Well, this Lie algebra was precisely produced to satisfy this, namely starting with the 2-generator free 5-step-nilpotent Lie algebra, observing that the degree 5 component splits as $\mathrm{GL}_2$-module as sum of a 4-dimensional and a 2-dimensional module, and killing the 4-dimensional module. The resulting Lie algebra is the free Lie algebra in the variety of 5-step-nilpotent Lie algebras satisfying the additional law $[x,[x,[x,[x,y]]]]=0$.)

Hence, all codimension-1 subalgebras of the above Lie algebra are isomorphic. Hence it is enough to see that a single one, say the one with basis $(e_i)_{2\le i\le 10}$ is 3-step nilpotent. Namely, its derived subalgebra has basis $(e_5,e_7,e_8,e_9,e_{10})$ and the next step in the central series has basis $(e_8,e_{10})$ and is central therein.

Now consider this Lie algebra over $\mathbf{Z}/p\mathbf{Z}$ with $p>5$. Then Lazard proved that the Baker–Campbell–Hausdorff formula produces a group law for the lower central series that coincides with the Lie-algebraic one, and subgroups also coincide with subalgebras. Thus the resulting group works.

Not always. Here is an answer in the realm of Lie algebras, and below I'll make it an answer in the realm of $p$-groups (a counterexample of order $p^{10}$ and class $5$ (and exponent $p$) in which every proper subgroup has class $\le 3$).

Fix an arbitrary ground field. Consider the vector space of basis $(e_i)_{1\le i\le 10}$. Make it a Lie algebra with the brackets (using the shortcut $i.j|x$ to mean $[e_i,e_j]=x=-[e_j,e_i]$, and other brackets being meant to be zero):

\begin{gather*} 1.2|e_3,\qquad 1.3|e_4,\;2.3|e_5,\qquad 1.4|e_6,\;1.5|e_7,\;2.4|e_7,\;2.5|e_8 \\ 1.7|e_9,\;2.7|-2e_9,\;1.8|e_{10},\;2.7|-2e_{10}. \end{gather*}

Then this satisfies Jacobi, and also satisfies that $[x,[x,[x,[x,y]]]]=0$ for all $x$, $y$ in the Lie algebra; the latter making use of the choice of $-2$ coefficients.

This is also a (Carnot) graded Lie algebra with $e_1$, $e_2$ of degree $1$, $e_3$ of degree $2$, $e_4$, $e_5$ of degree $3$, $e_6$, $e_7$, $e_8$ of degree $4$, and $e_9$, $e_{10}$ of degree $5$.

One then sees that the natural action of $\mathrm{GL}_2$ in the degree 1 space extends to the whole Lie algebra.

(Well, this Lie algebra was precisely produced to satisfy this, namely starting with the 2-generator free 5-step-nilpotent Lie algebra, observing that the degree 5 component splits as $\mathrm{GL}_2$-module as sum of a 4-dimensional and a 2-dimensional module, and killing the 4-dimensional module. The resulting Lie algebra is the free Lie algebra in the variety of 5-step-nilpotent Lie algebras satisfying the additional law $[x,[x,[x,[x,y]]]]=0$.)

Hence, all codimension-1 subalgebras of the above Lie algebra are isomorphic. Hence it is enough to see that a single one, say the one with basis $(e_i)_{2\le i\le 10}$ is 3-step nilpotent. Namely, its derived subalgebra has basis $(e_5,e_7,e_8,e_9,e_{10})$ and the next step in the central series has basis $(e_8,e_{10})$ and is central therein.

Now consider this Lie algebra over $\mathbf{Z}/p\mathbf{Z}$ with $p>5$. Then Lazard proved that the Baker–Campbell–Hausdorff formula produces a group law for which the lower central series coincides with the Lie-algebraic one, and subgroups also coincide with subalgebras. Thus the resulting group works.

Not always. Here is an answer in the realm of Lie algebras, and below I'll make it an answer in the realm of $p$-groups (a counterexample of order $p^{10}$ and class $5$ (and exponent $p$) in which every proper subgroup has class $\le 3$.

Fix an arbitrary ground field. Consider the vector space of basis $(e_i)_{1\le i\le 10}$. Make it a Lie algebra with the brackets (using the shortcut $i.j|x$ to mean $[e_i,e_j]=x=-[e_j,e_i]$, and other brackets being meant to be zero):

$$1.2|e_3,\qquad 1.3|e_4,\;2.3|e_5,\qquad 1.4|e_6,\;1.5|e_7,\;2.4|e_7,\;2.5|e_8$$ $$1.7|e_9,\;2.7|-2e_9,\;1.8|e_{10},\;2.7|-2e_{10}.$$

Then this satisfies Jacobi, and also satisfies that $[x,[x,[x,[x,y]]]]=0$ for all $x,y$ in the Lie algebra; the latter making use of the choice of $-2$ coefficients.

This is also a (Carnot) graded Lie algebra with $e_1,e_2$ of degree $1$, $e_3$ of degree $2$, $e_4,e_5$ of degree $3$, $e_6,e_7,e_8$ of degree $4$, and $e_9,e_{10}$ of degree $5$.

One then sees that the natural action of $\mathrm{GL}_2$ in the degree 1 space extends to the whole Lie algebra.

(Well, this Lie algebra was precisely produced to satisfy this, namely starting with the 2-generator free 5-step-nilpotent Lie algebra, observing that the degree 5 component splits as $\mathrm{GL}_2$-module as sum of a 4-dimensional and a 2-dimensional module, and killing the 4-dimensional module. The resulting Lie algebra is the free Lie algebra in the variety of 5-step-nilpotent Lie algebras satisfying the additional law $[x,[x,[x,[x,y]]]]=0$.)

Hence, all codimension-1 subalgebras of the above Lie algebra are isomorphic. Hence it is enough to see that a single one, say the one with basis $(e_i)_{2\le i\le 10}$ is 3-step nilpotent. Namely, its derived subalgebra has basis $(e_5,e_7,e_8,e_9,e_{10})$ and the next step in the central series has basis $(e_8,e_{10})$ and is central therein.

Now consider this Lie algebra over $\mathbf{Z}/p\mathbf{Z}$ with $p>5$. Then Lazard proved that the Baker-Campbell-Hausdorff formula produces a group law for the lower central series coincides with the Lie-algebraic one, and subgroups also coincide with subalgebras. Thus the resulting group works.

Not always. Here is an answer in the realm of Lie algebras, and below I'll make it an answer in the realm of $p$-groups (a counterexample of order $p^{10}$ and class $5$ (and exponent $p$) in which every proper subgroup has class $\le 3$).

Fix an arbitrary ground field. Consider the vector space of basis $(e_i)_{1\le i\le 10}$. Make it a Lie algebra with the brackets (using the shortcut $i.j|x$ to mean $[e_i,e_j]=x=-[e_j,e_i]$, and other brackets being meant to be zero):

\begin{gather*} 1.2|e_3,\qquad 1.3|e_4,\;2.3|e_5,\qquad 1.4|e_6,\;1.5|e_7,\;2.4|e_7,\;2.5|e_8 \\ 1.7|e_9,\;2.7|-2e_9,\;1.8|e_{10},\;2.7|-2e_{10}. \end{gather*}

Then this satisfies Jacobi, and also satisfies that $[x,[x,[x,[x,y]]]]=0$ for all $x$, $y$ in the Lie algebra; the latter making use of the choice of $-2$ coefficients.

This is also a (Carnot) graded Lie algebra with $e_1$, $e_2$ of degree $1$, $e_3$ of degree $2$, $e_4$, $e_5$ of degree $3$, $e_6$, $e_7$, $e_8$ of degree $4$, and $e_9$, $e_{10}$ of degree $5$.

One then sees that the natural action of $\mathrm{GL}_2$ in the degree 1 space extends to the whole Lie algebra.

(Well, this Lie algebra was precisely produced to satisfy this, namely starting with the 2-generator free 5-step-nilpotent Lie algebra, observing that the degree 5 component splits as $\mathrm{GL}_2$-module as sum of a 4-dimensional and a 2-dimensional module, and killing the 4-dimensional module. The resulting Lie algebra is the free Lie algebra in the variety of 5-step-nilpotent Lie algebras satisfying the additional law $[x,[x,[x,[x,y]]]]=0$.)

Hence, all codimension-1 subalgebras of the above Lie algebra are isomorphic. Hence it is enough to see that a single one, say the one with basis $(e_i)_{2\le i\le 10}$ is 3-step nilpotent. Namely, its derived subalgebra has basis $(e_5,e_7,e_8,e_9,e_{10})$ and the next step in the central series has basis $(e_8,e_{10})$ and is central therein.

Now consider this Lie algebra over $\mathbf{Z}/p\mathbf{Z}$ with $p>5$. Then Lazard proved that the Baker–Campbell–Hausdorff formula produces a group law for the lower central series that coincides with the Lie-algebraic one, and subgroups also coincide with subalgebras. Thus the resulting group works.