In their paper, Darmon, Diamond and Taylor remarked the following :
(the previous paragraph of Section 2.2 (p. 55), <https://www.math.wisc.edu/~boston/ddt.pdf>)


If $\rho : G \rightarrow GL\_2(\mathbb{C})$ is irreducible, then $\text{ad}^0 \rho$ is either irreducible or the direct sum of two representations (one is a non-trivial character of order 2 and the other is an induced representation.) For more detail, see the file linked above. Can someone help me to understand this remark?



Remark on Fermat's Last Theorem by Darmon, Diamond and Taylor
The assocated $\mathbb{C}G$-module, say $M$, is $2$-dimensional, and 
${\rm End}\_{\mathbb{C}}(M)$ is $4$-dimensional. The action of $g \in G$ on 
${\rm End}\_{\mathbb{C}}(M)$ is via conjugation by $\rho(g)$. Then ${\rm End}\_{\mathbb{C}}(M)$ decomposes as a direct sum of two modules under this this action as $S \oplus {\rm End}^{0}(M)$, where $S$ is the set of scalar matrices, and ${\rm End}^{0}(M)$ is the set of matrices of trace $0$. The latter module is $3$-dimensional. 


The representation $\rho$ may be primitive or imprimitive. It is primitive if it can't be induced from a representation of a proper subgroup, and imprimitive if it can be so induced.
In the latter case, it must be induced from a (non-trivial) $1$-dimensional representation of a subgroup $H$ of index $2$. This means that there are two $H$-invariant $1$-dimensional spaces $U$ and $V$ which are interchanged by $G$, and such that the actions of $H$ on $U$ and $V$ are different ( otherwise the induced representation of $G$ would not be irreducible).


In that case, with respect to the right basis, $G$ is acting by monomial matrices. The action of $G$ on ${\rm End}\_{\mathbb{C}}(M)$ is also by monomial matrices with respect to a suitable basis, since the Kronecker product of monomial matrices is monomial.


Note that, in this case, the fixed point subspace of $H$ on ${\rm End}\_{\mathbb{C}}(M)$ is $2$-dimensional. This fixed point space is $G$-invariant, and completely reducible since it is really a module for the finite group $G/H$. It therefore decomposes as the direct sum of the trivial module (from $S$, which is the whole space of $G$-fixed points by Schur's Lemma) and a $1$-dimensional module on which $H$ acts trivially and every element of $G \backslash H$ acts as $-1$.


In the monomial action of $G$ on ${\rm End}\_{\mathbb{C}}(M)$, one can see that there remains a $2$-dimensional summand on which $H$ acts as non-trivial diagonal matrices, and $G$ interchanges the corresponding $1$-dimensional subspaces. This is the remaining induced $2$-dimensional summand.


There is one scenario where that $2$-dimensional module is not irreducible for $G$ (and the authors did not specify whether that two dimensional module was irreducible or not so this remark is tangential). Let $\alpha$ and $\beta$ be the respective characters of $H$ afforded by the two $H$-invariant one-dimensional subspaces of $M$. Then the linear characters of $H$ in its diagonal action on ${\rm End}\_{\mathbb{C}}(M)$ are $1,1,\alpha\beta^{-1}$ and $\alpha^{-1}\beta$. Hence if $\alpha(h)^{2} = \beta(h)^{2}$
for all $h \in H$, then $H$ has the same action on the two remaining spaces, and $G$ does not act irreducibly on that space. This can happen if $H$ itself has a normal subgroup $K$ of index $2$, in which case $H$ has a linear character $\lambda$ with kernel $K$.
Then if we have $\beta = \lambda \alpha$, the corresponding induced 2-dimensional submodule of ${\rm End}\_{\mathbb{C}}(M)$ module is not irreducible for $G$, and decomposes as a direct sum of two one-dimensional modules. The dihedral group or order $8$ has a $2$-dimensional irreducible representation of this type, induced from a representation of a Klein $4$-subgroup.


Hence the second scenario specified by the authors occurs if the representation is imprimitive. If the representation is primitive, then it is easy to see (using Clifford's Theorem) that every normal subgroup of $G$ which does not act as scalars on $M$ acts irreducibly. If ${\rm End}\_{\mathbb{C}}^{0}$ were reducible, then it would have a non-trivial $1$-dimensional irreducible $G$-submodule whose kernel $H$ would be a proper normal subgroup of $G$. Then $H$ does not acts irreducibly on $M$ by Schur's Lemma, so by primitivity, $H$ acts as scalars on $M$ and $H \leq S(G)$. Now the derived group $G^{\prime}$ is contained in $H$ since $G/H$ is Abelian. Hence $G^{\prime}$ acts as scalars on $M$. In fact every element of $G^{\prime}$ acts as $\pm I$ on $M$, since an element of $G^{\prime}$ must act a matrix of determinant $1$. Now for any element $x \in G \backslash S(G)$ we see that $\langle \rho(x), -\rho(x) \rangle$ is an Abelian (proper) normal subgroup of ${\rm Im}(\rho)$ ( since it contains the derived group $[\rho(G),\rho(G)]$) which does not consist of scalars, in contradiction to the primitivity of $\rho$.


