Is there a non-split algebraic torus (over a finite field) satisfying the following properties? Is there a non-split algebraic torus $T$ (over a finite field $\mathbb{F}_{\!q}$) satisfying the following properties?

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*$T$ is not $\mathbb{F}_{\!q}$-isomorphic to the direct product of algebraic tori of smaller dimensions;

*The group $T(\mathbb{F}_{\!q})$ contains (or even isomorphic to) a large subgroup of the form $(\mathbb{Z}/n)^m$ for some naturals $n, m > 1$.

 A: $\def\FF{\mathbb{F}}\def\ZZ{\mathbb{Z}}\def\Id{\text{Id}}$The question is a little vague as to what large means, but I will show that, for any prime $\ell$ and any exponent $n$, the group $T(\FF_q)$ can contain $(\ZZ/\ell \ZZ)^n$. As Will Sawin writes in comments, this corresponds to constructing a matrix $M$ of finite order in $GL_d(\ZZ)$ such that $M$ is not decomposable as $M_1 \oplus M_2$ and such that $M-q \Id$ has an $n$-dimensional kernel modulo $\ell$.

Before I do the general case, an example to show that $T(\FF_q)$ can contain $(\ZZ/2 \ZZ)^2$. Let $\omega$ be a primitive cube root of unity.  Note that we have an injection
$$\ZZ[X]/(X^6-1) \ZZ[X] \hookrightarrow \ZZ \oplus \ZZ[\omega] \oplus \ZZ[\omega] \oplus \ZZ$$
by $X \mapsto (1,\omega,-\omega,-1)$. Let $A \subset \ZZ \oplus \ZZ[\omega] \oplus \ZZ[\omega] \oplus \ZZ$ be the subring
$$\{ (\alpha, \beta, \gamma, \delta) : \alpha \bmod 3 \equiv \beta \bmod 1-\omega,\ \beta \bmod 2 \equiv \gamma \bmod 2,\ \gamma \bmod 3 \equiv \delta \bmod 1-\omega \}$$
of $\ZZ \oplus \ZZ[\omega] \oplus \ZZ[\omega] \oplus \ZZ$.
Then $A$ is an indecomposable $\ZZ[X]/(X^6-1)\ZZ[X]$-module and, as a $\ZZ$-module, $A \cong \ZZ^6$. So the action of $X$ on $A$ gives a $6 \times 6$ integer matrix $M$ whose $6$-th power is $1$, and which can't be put in block diagonal form.
Now, let $q$ be any odd prime; we count solutions of $Mx = qx$ in $A/2A$. Of course, this is just the same as $Mx=x$.
We have
$$A/2A = \FF_2 \times (\text{nilpotent extension of $\FF_4$}) \times \FF_2$$
and $M$ acts on $A/2A$ by $(1,\omega, 1)$. The point is that the condition modulo $1-\omega$ disappears modulo $2$.
Then the $1$-eigenspace of $M$ is two dimensional.

Now, we need to do the general case. The notation is awful, I'm afraid.
For any positive integer $d$, let $\zeta_d$ be a primitive $d$-th root of unity. Then $\ZZ[X]/(X^r-1) \ZZ[X]$ injects into the ring $\prod_{d|r} \ZZ[\zeta_d]$. I'll write $(\alpha_d)_{d|r}$ for coordinates on the product $\ZZ[\zeta_d]$. Let $p^k s$ be a divisor of $r$ with $k>1$ and $p$ not dividing $s$. Then, on the image of $\ZZ[X]/(X^r-1) \ZZ[X]$, we have $\alpha_{p^{k-1} s} \bmod \langle 1-\zeta_{p^{k-1}}, p \rangle \equiv \alpha_{p^k s} \bmod \langle 1-\zeta_{p^{k}}, p \rangle$.
Let $E$ be any set of pairs of the form $(p^{k-1} s, p^k s)$ for $p^k s| r$ and  $p$ not dividing $s$. Let $A$ be the subring of $\prod_{d|r} \ZZ[\zeta_d]$ where we impose that $\alpha_{p^{k-1} s}  \bmod \langle p,1-\zeta_{p^{k-1}} \rangle \equiv \alpha_{p^k s} \bmod \langle p,1-\zeta_{p^k}\rangle$ for $(p^{k-1} s, p^k s) \in E$. Then $A$ is a $\ZZ[X]/(X^r-1) \ZZ[X]$-module. If the graph with edge set $E$ is connected, then $A$ is not a direct sum. So, for any such connected subgraph $E$, we get an $r \times r$ matrix with $M^r = 1$ which is not a direct sum.
In particular, take $r = p \ell^{n-1}$ for some auxilliary prime $p \neq \ell$. Let $E$ be the set of pairs of the form $(\ell^i, p \ell^i)$, $0 \leq i \leq n-1$ and $(p \ell^{j-1}, p \ell^j)$ for $1 \leq j \leq n-1$.
We want to understand $A/\ell A$. We can localize $p$ first, which make the conditions modulo $\langle p,1-\zeta_{p^k} \rangle$ vanish. So $A[p^{-1}] \cong \bigoplus_{j=0}^{n-1} \ZZ[\zeta_{\ell^j}] \oplus (\text{something else})$ and then $A/\ell A \cong A[p^{-1}]/\ell A[p^{-1}] \cong \bigoplus_{j=0}^{n-1} \ZZ[\zeta_{\ell^j}]/\ell \ZZ[\zeta_{\ell^j}] \oplus (\text{something else})/\ell(\text{something else})$.
Then each $\ZZ[\zeta_{\ell^j}]/\ell \ZZ[\zeta_{\ell^j}]$ gives a one dimension $1$-eigenspace for $X$, so the space of solutions to $Mx = x \bmod \ell$ is at least $n$-dimensional. Taking $q$ to be a prime $\equiv 1 \bmod \ell$, we see that $T(\FF_q)$ can contain $(\ZZ/\ell \ZZ)^n$.
