# Show that 2 representations are not equivalent and find all the irreducible representations of G.

Show that 2 representations are not equivalent and find all the irreducible representations of $G$.

The group $G=T_{16}$ has order 16 and presentation given by $G=\langle a,b : a^8=b^2=1, b^{-1}ab=a^{-1} \rangle.$
Take $n = e^{2πi/8}$. For $1 ≤ j ≤ 3$, there is a representation $p_j(a) : G → \operatorname{GL}(2,\mathbb{C})$ determined by

$$p_j(a)=\begin{bmatrix}n^j & 0 \\ 0 & n^{-j}\end{bmatrix}$$
and

$$p_j(b)=\begin{bmatrix}0 & 1 \\ n^{4j} & 0\end{bmatrix}$$

(b.) If $1≤j,k≤3$ and $j$ is not equal to $k$, show that $p_j$ and $p_k$ are not equivalent. Hint: You may wish to consider characters.

So I am confused because does this mean I consider $p_1(a)$ with $p_2(b)$, $p_3(b)$ and $p_2(a)$ with $p_1(b)$, $p_3(b)$ and $p_3(a)$ with $p_1(b)$ and $p_2(b)$?

Also, I know that the definition of two representations being equivalent is that, $\sigma$ and $p$ are equivalent if there exists $T$ such that $\sigma(g)=T^{-1}p(g)T$, but the only example of have is where the lecturer just plucks the matrix $T$ out of thin air, saying she knows it works because she wrote the question and this doesn't help me at all!

The hint says to look at the characters, so I assume this question is answered easiest by doing that, so I just calculate the trace of each matrix and if the traces are not equal then the representations are not equivalent? I would appreciate guidance on answering a similar question using the matrices approach though in case I am asked to do that in my exam.

(c.) Find all the irreducible representations of $G$, briefly explaining why you have discovered them all.
I haven't tried this part yet, but I thought I would ask it now and can get help in it when I have finished the previous part. Thanks!

• "so I just calculate the trace of each matrix and if the traces are not equal then the representations are not equivalent?" This is exactly right. – Dustan Levenstein May 26 '14 at 20:07
• “Does this mean I consider...?” No. Pick $j,k$ such as $j=1$, $k=2$. For every $x,y$ (with $0 \leq x < 8$ and $0 \leq y < 2$), you compare $p_j(a^x b^y)$ to $p_k(a^x b^y)$. Is there a matrix that conjugates the $p_j$ to the $p_k$? If there was it would take the traces of the $p_j$ to the traces of the $p_k$, but conjugation doesn't change the trace. – Jack Schmidt May 26 '14 at 20:09
• Your group $T_{16}$ as originally written had order 32. I added in the obvious relation to give it order 16, but now it is the dihedral group of order 16. You should check your question carefully to get the correct group. – Jack Schmidt May 26 '14 at 20:10
• In particular, $p_j$ does not define a representation of your group for $j=1,3$. – Jack Schmidt May 26 '14 at 20:13
• @JackSchmidt, my question is definitely as I had previously stated it; The group G = $T_{16}$ has order 16 and presentation $G$=⟨$a,b: a^8 =1,a^4 =b^2,b^{−1}ab=a^{−1}$⟩ – user150068 May 27 '14 at 6:48