Return to Answer

As mentioned in the comments, conjecturally almost all finite groups are $2$-step nilpotent $2$-groups, so conjecturally the answer to 1) and 3) are that the limits both exist and both equal $1$; that is, almost all finite groups have even order and almost all finite groups are solvable (even nilpotent). As numerical evidence for this, almost all of the first $50$ billion groups have order $1024$. The conjectural answer to 2) is then that if $m$ is a power of $2$ then the limit is equal to $1$ and otherwise if $m$ has a nontrivial odd divisor then the limit is equal to $0$.

It's worth knowing as context here that a result due to Higman and Sims states that asymptotically the number of $p$-groups of order $p^n$ is $p^{ \frac{2}{27} n^3 + O \left( n^{8/3} \right)}$. The lower bound comes from counting $2$-step nilpotent $p$-groups; you can see an analogous argument for nilpotent Lie algebras here. Thinking of this count as a function of the order $p^n$ it's not hard to check that it's maximized, if $p^n$ is bounded by some reasonably large $N$, by making $p$ as small as possible (equivalently, by making $n$ as large as possible), which is what singles out $p = 2$. It should be possible to write down a similar heuristic argument showing that the count of nilpotent groups (which are products of their Sylow subgroups) is dominated by groups of order $2^n$ also.

As mentioned in the comments, conjecturally almost all finite groups are $2$-step nilpotent $2$-groups, so conjecturally the answers to 1) and 3) are that the limits both exist and both equal $1$; that is, almost all finite groups have even order and almost all finite groups are solvable (even nilpotent). As numerical evidence for this, almost all of the first $50$ billion groups have order $1024$. The conjectural answer to 2) is then that if $m$ is a power of $2$ then the limit is equal to $1$ and otherwise if $m$ has a nontrivial odd divisor then the limit is equal to $0$.

It's worth knowing as context here that a result due to Higman and Sims states that asymptotically the number of $p$-groups of order $p^n$ is $p^{ \frac{2}{27} n^3 + O \left( n^{8/3} \right)}$. The lower bound comes from counting $2$-step nilpotent $p$-groups; you can see an analogous argument for nilpotent Lie algebras here. Thinking of this count as a function of the order $p^n$ it's not hard to check that it's maximized, if $p^n$ is bounded by some reasonably large $N$, by making $p$ as small as possible (equivalently, by making $n$ as large as possible), which is what singles out $p = 2$. It should be possible to write down a similar heuristic argument showing that the count of nilpotent groups (which are products of their Sylow subgroups) is dominated by groups of order $2^n$ also.

As mentioned in the comments, conjecturally almost all finite groups are $2$-step nilpotent $2$-groups, so conjecturally the answer to 1) and 3) are that the limits both exist and both equal $1$; that is, almost all finite groups have even order and almost all finite groups are solvable (even nilpotent). As numerical evidence for this, almost all of the first $50$ billion groups have order $1024$. The conjectural answer to 2) is then that if $m$ is a power of two then the limit is equal to $1$ and otherwise if $m$ contains a nontrivial odd divisor then the limit is equal to $0$.

It's worth knowing as context here that a result due to Higman and Sims states that asymptotically the number of $p$-groups of order $p^n$ is $p^{ \frac{2}{27} n^3 + O \left( n^{8/3} \right)}$. The lower bound comes from counting $2$-step nilpotent $p$-groups; you can see an analogous argument for nilpotent Lie algebras here. Thinking of this count as a function of the order $p^n$ it's not hard to check that it's maximized, if $p^n$ is bounded by some reasonably large $N$, by making $p$ as small as possible (equivalently, by making $n$ as large as possible), which is what singles out $p = 2$. It should be possible to write down a similar heuristic argument showing that the count of nilpotent groups (which are products of their Sylow subgroups) is dominated by groups of order $2^n$ also.

As mentioned in the comments, conjecturally almost all finite groups are $2$-step nilpotent $2$-groups, so conjecturally the answer to 1) and 3) are that the limits both exist and both equal $1$; that is, almost all finite groups have even order and almost all finite groups are solvable (even nilpotent). As numerical evidence for this, almost all of the first $50$ billion groups have order $1024$. The conjectural answer to 2) is then that if $m$ is a power of $2$ then the limit is equal to $1$ and otherwise if $m$ has a nontrivial odd divisor then the limit is equal to $0$.

It's worth knowing as context here that a result due to Higman and Sims states that asymptotically the number of $p$-groups of order $p^n$ is $p^{ \frac{2}{27} n^3 + O \left( n^{8/3} \right)}$. The lower bound comes from counting $2$-step nilpotent $p$-groups; you can see an analogous argument for nilpotent Lie algebras here. Thinking of this count as a function of the order $p^n$ it's not hard to check that it's maximized, if $p^n$ is bounded by some reasonably large $N$, by making $p$ as small as possible (equivalently, by making $n$ as large as possible), which is what singles out $p = 2$. It should be possible to write down a similar heuristic argument showing that the count of nilpotent groups (which are products of their Sylow subgroups) is dominated by groups of order $2^n$ also.

As mentioned in the comments, conjecturally almost all finite groups are $2$-step nilpotent $2$-groups, so conjecturally the answer to 1) and 3) are that the limits both exist and both equal $1$; that is, almost all finite groups have even order and almost all finite groups are solvable (even nilpotent). As numerical evidence for this, almost all of the first $50$ billion groups have order $1024$. The conjectural answer to 2) is then that if $m$ is a power of two then the limit is equal to $1$ and otherwise if $m$ contains a nontrivial odd divisor then the limit is equal to $0$.

It's worth knowing as context here that a result due to Higman and Sims states that asymptotically the number of $p$-groups of order $p^n$ is $p^{ \frac{2}{27} n^3 + O \left( n^{8/3} \right)}$. The lower bound comes from counting $2$-step nilpotent $p$-groups; you can see an analogous argument for nilpotent Lie algebras here. Thinking of this count as a function of the order $p^n$ it's not hard to check that it's maximized, if $p^n$ is required to lie in some interval $[1, N]$, by making $p$ as small as possible (equivalently, by making $n$ as large as possible), which is what singles out $p = 2$.

As mentioned in the comments, conjecturally almost all finite groups are $2$-step nilpotent $2$-groups, so conjecturally the answer to 1) and 3) are that the limits both exist and both equal $1$; that is, almost all finite groups have even order and almost all finite groups are solvable (even nilpotent). As numerical evidence for this, almost all of the first $50$ billion groups have order $1024$. The conjectural answer to 2) is then that if $m$ is a power of two then the limit is equal to $1$ and otherwise if $m$ contains a nontrivial odd divisor then the limit is equal to $0$.

It's worth knowing as context here that a result due to Higman and Sims states that asymptotically the number of $p$-groups of order $p^n$ is $p^{ \frac{2}{27} n^3 + O \left( n^{8/3} \right)}$. The lower bound comes from counting $2$-step nilpotent $p$-groups; you can see an analogous argument for nilpotent Lie algebras here. Thinking of this count as a function of the order $p^n$ it's not hard to check that it's maximized, if $p^n$ is bounded by some reasonably large $N$, by making $p$ as small as possible (equivalently, by making $n$ as large as possible), which is what singles out $p = 2$. It should be possible to write down a similar heuristic argument showing that the count of nilpotent groups (which are products of their Sylow subgroups) is dominated by groups of order $2^n$ also.