# Action inside semidirect products

When considering a group G which is the Semidirect product of $$\mathbb{Z}_3$$ and $$\mathbb{Z}_7 \times \mathbb{Z}_7$$, where $$\mathbb{Z}_7 \times \mathbb{Z}_7$$ is the normal subgroup, if z is a generator for the sylow 3-subgroup of G, then z induces an automorphism of $$\mathbb{Z}_7 \times \mathbb{Z}_7$$ via conjugation. That means, z induces an invertible linear transformation $$T$$ of $$\mathbb{Z}_7 \times \mathbb{Z}_7$$ such that $$T^3=I$$. Therefore minimal polynomial m(x) of $$T$$ divides $$x^3-1 = (x-1)(x^2+x+1)$$ and must be of degree atmost 2. How is it taken that the minimal polynomial is of degree atmost 2?

Not sure what you are looking for when you say "how is it taken?", but if the question is whether some such $$T$$ other than $$I$$ exists, the answer is yes. The easiest way to see this is that the group of linear transformations of $$\mathbb{Z}/(7) \times \mathbb{Z}/(7)$$ to itself is $$GL_2(7)$$, the set of invertible 2x2 matrices with entries from $$\mathbb(Z)/(7)$$. That means you have $$7^2-1$$ choices for the first row, and $$7^2-7$$ choices for the second row, so the order is $$(7-1)^27(7+1)$$, and 3 divides that, so $$GL_2(7)$$ has an element $$T$$ of order 3.

Also using standard linear algebra results, you can write such a matrix (which will have minimal polynomial $$x^2+x+1$$) by writing the negatives of the non-leading coefficients on the right-most column, and ones on the subdiagonal. In this case, that gives $$\pmatrix{0 & -1 \\ 1 &-1}$$ as one possibility for $$T$$. (Obviously, there is are many more examples of matrices of order $$3$$ here since the 3-Sylow subgroup of $$GL_2(7)$$ is not normal.)

Note that for the above matrix you can observe directly that $$T(ix+jy)=-jx+(i-j)y$$, or $$T(x^iy^j)=x^{-j}y^{i-j}$$ in your notation.

Also note that since the field with $$7$$ elements has primitive cube roots of unity, you can always find a eigenbasis. For example, for $$T$$ as above, $$x+3y$$ and $$x+5y$$ are eigenvectors with eigenvalues $$4$$ and $$2$$, respectively. Thus, it is similar to $$\pmatrix{4 & 0 \\ 0 &2}$$. Thus shows that all actions in the semidirect product are reducible. In fact, this means there are four dissimilar ways for $$\mathbb{Z}/(3)$$ to act on $$\mathbb{Z}/(7) \times \mathbb{Z}/(7)$$, giving rise to four distinct non-isomorphic semidirect products (one of which is the direct product) and given by the matrices (each along with its square) $$\pmatrix{4 & 0 \\ 0 &2}$$, $$\pmatrix{2 & 0 \\ 0 &2}$$, $$\pmatrix{2 & 0 \\ 0 &1}$$, and $$\pmatrix{1 & 0 \\ 0 &1}$$. In other words, once you choose a basis that makes the action diagonal, $$\mathbb{Z}/(3)$$ can act on both, one, or neither of the basis elements, and if it acts on both it can act the differently or the same on each. (If it acts the same on each, we say it acts by homotheties.)

Note that if we had been considering instead semidirect products of $$\mathbb{Z}/(11) \times \mathbb{Z}/(11)$$ with $$\mathbb{Z}/(3)$$, we would not have been able to diagonalize, since $$\mathbb{Z}/(11)$$ has no non-trivial cube roots of unity.

• Thanks. Another small question. Suppose {x,y} is a basis for $\mathbb{Z}_7 \times \mathbb{Z}_7$. If $x = T(y) = z^{-1}yz$ , then can we simplify $T(y^i x^j) = z^{-1} y^i x^j z = y^{-j} x^{i-j}$ ?. I came across this simplification in a certain paper. The answer was taken as $y^{-j} x^{i-j}$, by taking $T(y^i x^j)$. I don't want to know how $T(y^i x^j)$ was taken. But I just need to know whether we can simplify $T(y^i x^j)$ to get the above answer with -j powers. – Buddhini Angelika Oct 6 '18 at 4:20
• Can someone explain the steps to come from $T(y^i x^j)$ to $y^{-j} x^{i-j}$? Assuming that all the other given details are correct. – Buddhini Angelika Oct 6 '18 at 4:24
• See the revised answer. – C Monsour Oct 6 '18 at 12:56
• Thanks. I need to clarify a little bit more about how you observed that $T(ix+jy)=-jx+(i-j)y$. Did you think $x=(1,0),y=(0,1)$ as standard basis ? Or is there a general way? – Buddhini Angelika Oct 8 '18 at 16:15