Let $G$ be a finite group and $\mathbb{F}_q$ be a finite field, where $q$ is a prime power. Let $\mathbb{F}_qG$ be a finite semisimple group algebra.
Then Wedderburn decomposition theorem implies that $\mathbb{F}_qG$ is direct sum of matrix rings over division rings, i.e.,$$\mathbb{F}_qG\cong \oplus_{i=1}^t M(n_i, \mathbb{F}_{q_i}).$$
Here $\mathbb{F}_{q_i}$ is a finite extension of $\mathbb{F}_{q}$, $n_i\geq 1$ and $t\geq 1$ are positive integers.
Given: Suppose that $\mathbb{F}_{q}$ is such that $$\mathbb{F}_qG\cong \oplus_{i=1}^t M(n_i, \mathbb{F}_{q}).$$ This means $q_i=q$ for each $i$.
Let $q'\neq q$ be another prime power such that the group algebra $\mathbb{F}_{q'}G$ is again semisimple.
Suppose that $M\left(n, \mathbb{F}_{(q')^r}\right)$ is a Wedderburn component of the group algebra $\mathbb{F}_{q'}G$ for some $r\geq 2$ and any positive integer $n$.
Then I want to show that $M\left(n, \mathbb{F}_{q}\right)$ must be a Wedderburn component of the group algebra $\mathbb{F}_qG$.
I have tried various examples and found this result true. Please help me to prove this.