Infinite group with elements of finite order It is easy to construct an infinite group with all elements of finite order, e.g., the infinite direct sum of $\mathbb{Z}/2\mathbb{Z}$.
However, I wonder that for a given integer $n>2$, if there exists an infinite group that satisfies:

*

*For each $1\le k\le n$, there exists an element of order $k$,

*For each $k>n$ (including $k=\infty$), there does not exist an element of order $k$.

If there is no group satisfying the restriction, we can consider replacing condition 2 with
2.1 For each $k>n$ (excluding $k=\infty$), there does not exist an element of order $k$.
(Update: 1+(2.1) can be satisfied by $C_2 * C_3 * \ldots * C_n$, where $*$ is the free product. Thanks to @user10354138)
Or consider some special $n$, e.g., a prime $p$, or a factorial number $k!$.
I have no clue to solve this problem using the direct sum. For example, for $n=3$, we cannot use $(\mathbb{Z}/2\mathbb{Z})^\infty \times (\mathbb{Z}/3\mathbb{Z})$ because there is an element of order $6$.
Any help is appreciated!

For $n=2$ we can take $G$ to be a subdirect product of  infinitely many copies of $A_4$, where $G$ has an infinite normal elementary abelian $2$-subgroup $N$ with $|G/N|=3$ and $C_G(N)=N$.
For $n=4$, do a similar construction with $S_4$, with $G/N = S_3$.
For  $n=5$, there is a similar example with $G/N = A_5$, where (as $A_5$-module) $N$ is a direct sum of infinitely many copies of the $2$-dimensional module over ${\mathbb F}_4$ (i.e. the natural module for ${\rm SL}(2,4) \cong A_5$).
For $n=6$ we can just take the direct product of the $n=4$ example with $C_2$.
I don't know about higher $n$, but it is known that the only finite group with element orders $\{1,2,3,4,5,6,7\}$ is $A_7$, so it is perhaps unlikely
that infinite examples exist.
