I don’t understand some of the following text (from Adkins’ Algebra):

1- Why if qq does not divide p−1p-1 then ϕh=1N\phi_h = 1_N (i.e. HH is normal) ? Exercise 11 doesn’t help since it says “… if nh=hnnh=hn …” and not conclude that!

2- If q|p−1q|p-1 why there are nontrivial homomorphisms, especially when in both cases ϕ:Zq→Aut(N)≅Z∗p\phi : Z_q \to Aut(N) \cong Z^*_p?

3- Why b−1abb^{-1}ab have to be in a form of ara^r?! And why rq≡1 mod pr^q \equiv 1 \ mod \ p?

Thank you.

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1 Answer

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(1) There can’t be a nontrivial homomorphism G→HG\to H if |G||G| and |H||H| are coprime. This follows from the first isomorphism theorem (the image is isomorphic to a quotient of GG) and Lagrange’s theorem (which implies quotients of GG have size dividing |G||G|). In particular, if qq and p−1p-1 are coprime (equivalently q∤p−1q\nmid p-1), then the homomorphism Zq→Aut(Zp)\mathbb{Z}_q\to\mathrm{Aut}(\mathbb{Z}_p) must be trivial, since the order of Aut(Zp)\mathrm{Aut}(\mathbb{Z}_p) is just p−1p-1.

(2) If q∣p−1q\mid p-1 then Aut(Zp)\mathrm{Aut}(\mathbb{Z}_p) must have an element of order qq, and then there is a homomorphism which maps Zq\mathbb{Z}_q’s generator to that element of order qq. This will be a nontrivial homomorphism.

(3) If N=⟨a⟩N=\langle a\rangle is normal then bab−1∈⟨a⟩bab^{-1}\in\langle a\rangle. Since all elements of ⟨a⟩\langle a\rangle are of the form ara^r for some exponent rr, that means bab−1=arbab^{-1}=a^r for some rr. (I don’t know why the text is writing the conjugation backwards.) Notice that bab−1bab^{-1} is none other than ϕb(a)\phi_b(a). Applying ϕb\phi_b over and over again is tantamount to conjugating by powers of bb. That is,

ϕb∘⋯∘ϕb⏟n(a)=b(⋯bab−1⋯)b−1=bnab−n=ϕbn(a).\underbrace{\phi_b\circ\cdots\circ\phi_b}_n(a)=b(\cdots bab^{-1}\cdots)b^{-1}=b^nab^{-n}=\phi_{b^{\large n}}(a).

(Of course, ϕnb=ϕbn\phi_b^n=\phi_{b^{\large n}} is just a corollary to the fact that x↦ϕxx\mapsto \phi_x is a group homomorphism.)

But notice that, if ϕb(a)=ar\phi_b(a)=a^r and ϕb\phi_b is itself a group homomorphism, then

ϕb(ϕb(a))=ϕb(ar)=ϕb(a)r=(ar)r=ar2,ϕb(ϕb(ϕb(a)))=ϕb(ar2)=ϕb(a)r2=(ar)r2=ar3,⋮ϕb∘⋯∘ϕb⏟n(a)=ϕb(arn−1)=ϕb(a)rn−1=(ar)rn−1=arn\begin{array}{llll} \phi_b(\phi_b(a)) & = & \phi_b(a^r) & = & \phi_b(a)^r & = &(a^r)^r & = & a^{r^{\large 2}}, \\

\phi_b(\phi_b(\phi_b(a))) & = & \phi_b(a^{r^{\large 2}}) & = & \phi_b(a)^{r^{\large 2}} & = & (a^r)^{r^{\large 2}} & = & a^{r^{\large 3}}, \\

& & & \vdots \\ \underbrace{\phi_b\circ\cdots\circ\phi_b}_n(a) & = & \phi_b(a^{r^{\large n-1}}) & = & \phi_b(a)^{r^{\large n-1}} & = & (a^r)^{r^{\large n-1}} & = & a^{r^{\large n}} \end{array}

In particular, this means ϕbq(a)=arq\phi_{b^q}(a)=a^{r^{\large q}}. But ϕbq(a)=ϕe(a)=a\phi_{b^q}(a)=\phi_e(a)=a. Therefore

arq=a⇒arq−1=e⇒p∣(rq−1)i.e.rq≡1modp.a^{r^{\large q}}=a \quad\Rightarrow\quad a^{r^{\large q}-1}=e \quad \Rightarrow \quad p\mid (r^q-1) \quad\textrm{i.e.}\quad r^q\equiv 1\bmod{p}.

(3) is very clear. But I don’t understand (1) and (2), esp. (1). H/ker(د•) \cong Im(د•)H/ker(د•) \cong Im(د•). Then I understand (1) if Im(د•)Im(د•) is either surjective or a subgroup of Aut(Z_p)Aut(Z_p). Is it the case?

– Liebe

Oct 21 at 3:52

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@Liebe If \phi:G\to H\phi:G\to H is a group homomorphism them \mathrm{Im}(\phi)\mathrm{Im}(\phi) is a subgroup of HH, yes.

– arctic tern

Oct 21 at 3:54