I'm teaching myself abstract algebra using Abstract Algebra by Dummit&Foote. I have trouble understanding the last sentence of the second paragraph. Can someone elaborate on the example given in the brackets? Why does such subgroup exist?
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2Does this answer your question? Group of even order contains an element of order 2 It proves that "the quotient by the subgroup ${\pm1}$ contains a subgroup of order $2$". – Anne Bauval Dec 31 '22 at 07:43
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1Please do not rely on pictures of text. – Shaun Dec 31 '22 at 13:36
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@AnneBauval but how do we know that the preimage is a subgroup of order 4? – Irene Jan 01 '23 at 20:01
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1For a surjective morphism, the order of the preimage of a subgroup N is the order of the kernel times the order of N. This proof is much simpler than the one in the answer you accepted. – Anne Bauval Jan 01 '23 at 20:52
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1@AnneBauval Thanks! This proof makes more sense to me – Irene Jan 01 '23 at 21:16
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From the reasoning given in the proof:
Consider the homomorphism(?) $\varphi:(\mathbb Z_p )^\times\rightarrow \frac{(\mathbb Z_p )^\times}{\{\pm 1\}},\varphi(x)=x^2$
Note:$\left|\frac{(\mathbb Z_p )^\times}{\{\pm 1\}}\right|=\frac{p-1}{2}$ but given $4|(p-1)\implies2|\frac{p-1}{2}$,hence by cauchy's theorem $\frac{(\mathbb Z_p )^\times}{\{\pm 1\}}$ has a subgroup of order $2$ say $H$.
Now we have the subgroup $\varphi^{-1}(H)$ of $(\mathbb Z_p )^\times$, which has order of form $2^n$ with $n\geq 2$ (since the map $\varphi$ takes an element of odd prime order p to an element of odd prime order p ). Also we have that $\varphi^{-1}(H)$ must contain an element of order $4$, for if all non-trivial elements are of order $2$, then all will be annihilated by the map $\varphi$, and contradicting that $H$ is non trivial. Hence we have a (cyclic) subgroup of order $4$ generated by the above element.(In fact $\varphi^{-1}(H)$ itself is this group,which is evident from the fact proved above in the theorem that there is a unique element of order 2, and also noting that if $x$ is an element of order $2^m$,then order of $x^2$ is $2^{m-1}$,$m\geq1$)
Which is the required subgroup.
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Sorry, I jumped the gun. I didn't quite realize that you were discussing the converse part. Anyway, there is a simpler proof for the case $p=2$ of Cauchy. Anyway, I'm sorry, I need more coffee. – Jyrki Lahtonen Dec 31 '22 at 07:39
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hmm...But how do we know that this map is surjective? If the map is not surjective, then it is possible that the image of $\varphi^{-1}(H)$ is trivial, which is not equal to $H$. – Irene Dec 31 '22 at 19:30
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Well, since we're assuming $4\mid p-1$, why not just note that $(\Bbb Z/p\Bbb Z)^×\cong\Bbb Z/(p-1)\Bbb Z$.
And cyclic groups have subgroups of each order dividing the order of the group.
But it appears the author may not want to assume the cyclicity of the group of units here.
Abelian groups also have subgroups of every order dividing the group's order. They're called CLT (converse to Lagrange's theorem) groups. So we get the result.
The authors seem to have taken an extra step. Did you have a result about Abelian groups of even order having an element of order $2$? Idk, because I don't have the book. Come to think of it, you probably have Cauchy's theorem.
So, as for the last sentence, when you pullback (or this is where you want the correspondence theorem) a subgroup of order $2$ under a homomorphism with kernel of order two, you get a subgroup $H\le(\Bbb Z/p\Bbb Z)^×$ of order $4$.
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5It is morally certain that this theorem appears well before they have proved that $(\Bbb{Z}/p\Bbb{Z})^*$ is cyclic. – Jyrki Lahtonen Dec 31 '22 at 07:35
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@JyrkiLahtonen In both the number theory and abstract algebra courses I've taken, we proved that $(\mathbb{Z}/p \mathbb{Z})^*$ is cyclic before this result. – legionwhale Dec 31 '22 at 12:20