I want to solve the following exercise from Dummit & Foote's Abstract Algebra text (p. 185 Exercise 15):

Let $p$ be an odd prime. Prove that every element of order $2$ in $GL_2(\mathbb{F}_p)$ is conjugate to a diagonal matrix with $\pm 1$ on the diagonal. Classify the groups of order $2p^2$. [If $A$ is a $2 \times 2$ matrix with $A^2=I$ and $v_1,v_2$ is a basis for the underlying vector space, look at $A$ acting on the vectors $w_1=v_1+v_2$ and $w_2=v_1-v_2$.]

I can't see how to use the hint in the brackets. However, assuming that the conjugacy classes in $GL_2(\mathbb{F}_p)$ of elements of order two are $\{\text{diag}(1,-1) \}=\{\text{diag}(-1,1) \},\{\text{diag}(-1,-1) \}$, I try to classify the groups of order $2p^2$ as follows:

There are two abelian groups of order $2p^2$: $Z_{p^2} \times Z_2$ and $Z_p \times Z_p \times Z_2$.

Let $G$ be a non-abelian group with $|G|=2p^2$ ($p$ is odd). Let $H \leq G$ be a Sylow $p$-subgroup of $G$. Since $|G:H|=2$, $H$ is normal in $G$. Also, the possible isomorphism types of $H$ are $Z_{p^2}$ and $Z_p \times Z_p$.

Let $K=\langle x \rangle \leq G$ be a Sylow 2-subgroup. Since $G=HK$ and $H \cap K=1$ we have $G \cong H \rtimes_\varphi K$ for some (non-trivial) homomorphism $\varphi:K \to \text{Aut}(H)$.

  • We begin with the case where $H \cong Z_{p^2}$ is cyclic: In that case $\text{Aut}(H) \cong Z_{p^2-p}$ is cyclic, hence admits a unique subgroup $\langle \sigma \rangle$ of order 2. The only non-trivial homomorphism in this case is defined by $\varphi(x)=\sigma$.

  • Now we turn to the case where $H \cong Z_p \times Z_p$ is elementary abelian: In this case $\text{Aut}(H) \cong GL_2(\mathbb{F}_p)$, and the discussion above makes room for at most two non-isomorphic semi-direct products. This depends on whether $\varphi(x)=\text{diag}(1,-1)$ or $\text{diag}(-1,-1)$.

I'd like some help with the following:

  1. How can one use the hint in the brackets in order to find the conjugacy classes of elements of order 2?

  2. How can I show that the last two semi-direct products are indeed not isomorphic to each other?



Suppose that $A$ is not diagonal. Then $$Av_1=av_1+bv_2$$ and $$Av_2=cv_1+dv_2$$ for some elements $a,b,c,d$, with one of $b,c\neq 0$. Thus we have $$A(v_1-v_2)=(a-c)v_1+(b-d)v_2$$ and $$A(v_1+v_2)=(a+c)v_1+(b+d)v_2$$ Thus $$v_1-v_2=(a-c)(av_1+bv_2)+(b-d)(cv_1+dv_2)=(a(a-c)+c(b-d))v_1+(b(a-c)+d(b-d))v_2$$ and $$v_1+v_2=(a+c)(av_1+bv_2)+(b+d)(cv_1+dv_2)=(a(a+c)+c(b+d))v_1+(b(a+c)+d(b+d))v_2$$ Thus $$ab+db+(bc+d^2)=1$$ $$ab+db-(bc+d^2)=-1$$ $$a^2+bc+(ac+cd)=1$$ $$a^2+bc-(ac+cd)=1$$ Thus $$b(a+d)=c(a+d)=0$$ and $$a^2+bc=1$$ Since $A$ is not diagonal, at least one of $b$ or $c$ is not $0$, hence $d=-a$ regardless of the basis we chose. Since $(I+A)(I-A)=0$, either $A=I$, $A=-I$, or $I+A$ and $I-A$ are both nonzero and not invertible. Thus $A$ has an eigenvector, so $A$ is conjugate to an upper triangular matrix. If this upper triangular matrix is not diagonal, then $a^2=1$ and the determinant is $-1$ (since it is equal to $-a^2-bc$). Let $$A=\left(\begin{array}{cc}\pm 1&b\\0&\mp 1\end{array}\right)$$ We have $$A^2=\left(\begin{array}{cc}1&0\\\mp b& 1\end{array}\right)$$ Since this is in fact the identity, $b=0$ and the conjugacy result follows.

To see that the two groups are not isomorphic, suppose $A$ has the automorphism as $-I$ and $B$ does not. Note that an isomorphism $A\to B$ would have to send the unique $p$-subgroup to the unique $p$-subgroup and also send the $2$-subgroup to some conjugate of $K$. Via an inner automorphism we may assume that the subgroup corresponding to $K$ is mapped to $K$. This means there is some basis under which $\varphi$ acts as $-I$ and some basis under which $\varphi$ acts as $\mathrm{diag}(1,-1)$. However, this is impossible as the two matrices are not conjugate. Thus $A$ and $B$ are not isomorphic.

  • $\begingroup$ Why is $A$ conjugate to an upper triangular matrix? $\endgroup$
    – user1337
    Feb 4 '15 at 5:07
  • 1
    $\begingroup$ @user1337 I made it explicit. $\endgroup$ Feb 4 '15 at 6:58
  • 1
    $\begingroup$ @user1337 I just realized that using $I-A^2=0$ makes the hint useless because it actually shows that $A$ has a basis of eigenvectors. $\endgroup$ Feb 4 '15 at 7:36
  • 1
    $\begingroup$ Thanks! That's a great observation $\endgroup$
    – user1337
    Feb 4 '15 at 10:54
  • $\begingroup$ How it follows that if the upper triangular matrix is not diagonal $a^2=1$? $\endgroup$
    – budi
    Aug 29 '17 at 13:09

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