MHB The group is isomorphic to one of the groups

[mathmari]

Hey!

I want to show that if $G$ is of order $2p$ with $p$ a prime, then $G\cong \mathbb{Z}_{2p}$ or $G\cong D_p$. I have done the following:

We have that $|G|=2p$, so there are $2$-Sylow and $p$-Sylow in $G$.
$$P\in \text{Syl}_p(G) , \ |P|=p \\ Q\in \text{Syl}_2(G) , \ |Q|=2$$

Let $x\in P$ and $y\in Q$, then $x^p=e$ and $y^2=e$.

Since the order of $x$ is $p$ and the order of $y$ is $2$, the subgroup that they generate has an order that divides the order of $G$, so it must be equal to the order of $G$.
Therefore, $G$ is generated by $x, y$.

Since $[G:\langle x\rangle ]=\frac{|G|}{|\langle x\rangle |}=\frac{2p}{p}=2$ we have that $\langle x\rangle \trianglelefteq G$.

Since $\langle x\rangle$ is a normal subgroup of $G$ we have that $$y^{-1}xy\in \langle x\rangle, \text{ where } y\in G \\ \Rightarrow y^{-1}xy=x^t, \text{ for some } 1\leq t\leq p-1$$

We have that $$x=e^{-1}xe=(y^2)^{-1}xy^2=y^{-1}(y^{-1}xy)y=y^{-1}x^ty=y^{-1}xyy^{-1}xy\cdots y^{-1}xy=(x^t)^t=x^{t^2}$$

So $x=x^{t^2} \Rightarrow e=x^{t^2-1}$.

Therefore, $p$ divides $t^2-1=(t-1)(t+1)$, so $p$ divides $t-1$ or $t+1$.

$$p\mid t-1 \Rightarrow t-1=0 \Rightarrow t=1 \\ p\mid t+1 \Rightarrow p=t+1 \Rightarrow t=p-1$$

Is this correct? (Wondering) When $t=1$ we have that $xy=yx$.
We want to show that $xy$ has order $2p$, so that the group is generated by $xy$, i.e. so that the group is cyclic, right? (Wondering)
We have that $(xy)^{2p} \overset{ xy=yx }{ = } x^{2p}y^{2p}=(x^p)^2(y^2)^p=e$.
Suppose that $2p$ is not the smallest power, then say $n$ is the smallest power such that $(xy)^n=e$. That means that $$n\mid 2p \Rightarrow n\mid 2 \text{ or } n\mid p \Rightarrow n\in \{1,2,p\}$$
If $n=1$, then $xy=e \Rightarrow x=y^{-1}=y$. Since the order of $y$ is $2$, we would have that the order $x$ is also $2$. That is not true, since the order $x$ is $p$. Or can $p$ take the value $2$ ? (Wondering)
If $n=2$ we have that $(xy)^2=e \Rightarrow x^2y^2=e \Rightarrow x^2=e$. That is only true when the order of $x$ is $2$, so when $p=2$.
If $n=p$ we have that $(xy)^p=e \Rightarrow x^py^p=e \Rightarrow y^p=e$. That is only true when the order of $y$ is $p$, so when $p=2$.

Is this correct? (Wondering) When $t=p−1$, we have that $x^p=e=y^2$ and $y^{−1}xy=x^{p−1}\Rightarrow yxy=x^{−1}$. So, $x$ and $y$ generate the dihedral group $D_p$, right? (Wondering)

 

[Deveno]

Hey!

I want to show that if $G$ is of order $2p$ with $p$ a prime, then $G\cong \mathbb{Z}_{2p}$ or $G\cong D_p$. I have done the following:

We have that $|G|=2p$, so there are $2$-Sylow and $p$-Sylow in $G$.
$$P\in \text{Syl}_p(G) , \ |P|=p \\ Q\in \text{Syl}_2(G) , \ |Q|=2$$

Let $x\in P$ and $y\in Q$, then $x^p=e$ and $y^2=e$.

Since the order of $x$ is $p$ and the order of $y$ is $2$, the subgroup that they generate has an order that divides the order of $G$, so it must be equal to the order of $G$.
Therefore, $G$ is generated by $x, y$.

Since $[G:\langle x\rangle ]=\frac{|G|}{|\langle x\rangle |}=\frac{2p}{p}=2$ we have that $\langle x\rangle \trianglelefteq G$.

Since $\langle x\rangle$ is a normal subgroup of $G$ we have that $$y^{-1}xy\in \langle x\rangle, \text{ where } y\in G \\ \Rightarrow y^{-1}xy=x^t, \text{ for some } 1\leq t\leq p-1$$

We have that $$x=e^{-1}xe=(y^2)^{-1}xy^2=y^{-1}(y^{-1}xy)y=y^{-1}x^ty=y^{-1}xyy^{-1}xy\cdots y^{-1}xy=(x^t)^t=x^{t^2}$$

So $x=x^{t^2} \Rightarrow e=x^{t^2-1}$.

Therefore, $p$ divides $t^2-1=(t-1)(t+1)$, so $p$ divides $t-1$ or $t+1$.

$$p\mid t-1 \Rightarrow t-1=0 \Rightarrow t=1 \\ p\mid t+1 \Rightarrow p=t+1 \Rightarrow t=p-1$$

Is this correct? (Wondering) When $t=1$ we have that $xy=yx$.
We want to show that $xy$ has order $2p$, so that the group is generated by $xy$, i.e. so that the group is cyclic, right? (Wondering)
We have that $(xy)^{2p} \overset{ xy=yx }{ = } x^{2p}y^{2p}=(x^p)^2(y^2)^p=e$.
Suppose that $2p$ is not the smallest power, then say $n$ is the smallest power such that $(xy)^n=e$. That means that $$n\mid 2p \Rightarrow n\mid 2 \text{ or } n\mid p \Rightarrow n\in \{1,2,p\}$$
If $n=1$, then $xy=e \Rightarrow x=y^{-1}=y$. Since the order of $y$ is $2$, we would have that the order $x$ is also $2$. That is not true, since the order $x$ is $p$. Or can $p$ take the value $2$ ? (Wondering)
If $n=2$ we have that $(xy)^2=e \Rightarrow x^2y^2=e \Rightarrow x^2=e$. That is only true when the order of $x$ is $2$, so when $p=2$.
If $n=p$ we have that $(xy)^p=e \Rightarrow x^py^p=e \Rightarrow y^p=e$. That is only true when the order of $y$ is $p$, so when $p=2$.

Is this correct? (Wondering) When $t=p−1$, we have that $x^p=e=y^2$ and $y^{−1}xy=x^{p−1}\Rightarrow yxy=x^{−1}$. So, $x$ and $y$ generate the dihedral group $D_p$, right? (Wondering)

What you have looks correct, but you can simplify it a bit.

First of all, you should recognize straight off the bat that $p = 2$ is a special case, and that treating it separately is beneficial. For groups of order $4$, the proposition is still true, as there are only two isomorphism classes:

$\Bbb Z_4$ and $\Bbb Z_2 \times \Bbb Z_2$ (the latter is sometimes considered a dihedral group).

As soon as you establish that either:

1) $yxy = x$ (so that $G$ is abelian), if $p$ is an ODD prime, it is immediate that the order of $xy$ is $\text{lcm}(2,p) = 2p$.

-OR-

2) $yxy = x^{p-1} = x^{-1}$, then $G$ is (isomorphic to) the dihedral group of order $2p$.

(I urge you to consider what happens if $x$ and $y$ are not distinct, perhaps this will make it clear why $p = 2$ should be handled separately).

 

[mathmari]

First of all, you should recognize straight off the bat that $p = 2$ is a special case, and that treating it separately is beneficial. For groups of order $4$, the proposition is still true, as there are only two isomorphism classes:

$\Bbb Z_4$ and $\Bbb Z_2 \times \Bbb Z_2$ (the latter is sometimes considered a dihedral group).

Is it known that there are only these two or do we have to prove it? (Wondering)

As soon as you establish that either:

1) $yxy = x$ (so that $G$ is abelian), if $p$ is an ODD prime, it is immediate that the order of $xy$ is $\text{lcm}(2,p) = 2p$.

-OR-

2) $yxy = x^{p-1} = x^{-1}$, then $G$ is (isomorphic to) the dihedral group of order $2p$.

(I urge you to consider what happens if $x$ and $y$ are not distinct, perhaps this will make it clear why $p = 2$ should be handled separately).

If $p$ is an odd prime we have the following:

We have that $|G|=2p$, so there are $2$-Sylow and $p$-Sylow in $G$.
$$P\in \text{Syl}_p(G) , \ |P|=p \\ Q\in \text{Syl}_2(G) , \ |Q|=2$$

Let $x\in P$ and $y\in Q$, then $x^p=e$ and $y^2=e$.

Since the order of $x$ is $p$ and the order of $y$ is $2$, the subgroup that they generate has an order that divides the order of $G$, so it must be equal to the order of $G$.
Therefore, $G$ is generated by $x, y$.

Since $[G:\langle x\rangle ]=\frac{|G|}{|\langle x\rangle |}=\frac{2p}{p}=2$ we have that $\langle x\rangle \trianglelefteq G$.

Since $\langle x\rangle$ is a normal subgroup of $G$ we have that $$y^{-1}xy\in \langle x\rangle, \text{ where } y\in G \\ \Rightarrow y^{-1}xy=x^t, \text{ for some } 1\leq t\leq p-1$$

We have that $$x=e^{-1}xe=(y^2)^{-1}xy^2=y^{-1}(y^{-1}xy)y=y^{-1}x^ty=y^{-1}xyy^{-1}xy\cdots y^{-1}xy=(x^t)^t=x^{t^2}$$

So $x=x^{t^2} \Rightarrow e=x^{t^2-1}$.

Therefore, $p$ divides $t^2-1=(t-1)(t+1)$, so $p$ divides $t-1$ or $t+1$.

$$p\mid t-1 \Rightarrow t-1=0 \Rightarrow t=1 \\ p\mid t+1 \Rightarrow p=t+1 \Rightarrow t=p-1$$

• When $t=1$, we have that $xy=yx$.
  We have that that the order of $xy$ is $\text{lcm}(2,p) = 2p$.
  Therefore, we have a cyclic abelian group of order $2p$. So, it is isomorphic to the cyclic group $\mathbb{Z}_{2p}$, right? (Wondering)
  
  Does it always hold that when the order of $x$ is $m$ and the order of $y$ is $n$ then the order of $xy$ is $\text{lcm}(m,n)$ ? (Wondering)
• When $t=p−1$, we have that $x^p=e=y^2$ and $y^{−1}xy=x^{p−1}\Rightarrow yxy=x^{−1}$. So, $x$ and $y$ generate the dihedral group $D_p$, right? (Wondering)

Is the case when $p$ is an odd prime correct? (Wondering)

 

[Deveno]

Is it known that there are only these two or do we have to prove it? (Wondering)

It's not hard to prove. A "high-level" proof goes something like this:

Let $|G| = 4$.

$4 = p^2$, for the prime $p = 2$. Therefore, any group of order $4$ is abelian. For if not, then, being a $p$-group, it has a non-trivial center. But then $G/Z$ has order 2, and is cyclic, and thus G is abelian, contradiction.

By the structure theorem for finitely-generated abelian groups:

$G \cong \Bbb Z_4$, or $G \cong \Bbb Z_2 \times \Bbb Z_2$, as the only two ways to write 4 as a product of powers of two are:

$4 = 2^2$
$4 = 2\cdot 2$

(This "high-level" proof works for ANY prime $p$, showing there are just two non-isomorphic groups of order $p^2$).

A "low-level" proof goes something like this: Let $G = \{e,a,b,c\}$. The possible orders for $a$ are 2, and 4. If the order of $a$ is 4, $G$ is cyclic, and isomorphic to $\Bbb Z_4$. A similar argument holds if the order of $b$ is 4.

So assume that $a^2 = b^2 = e$. Can $G$ still be cyclic? No, because a cyclic group of order 4 (with generator $x$) has two elements of order 4, $x$ and $x^{-1} = x^3$. So it must be that $c^2 = e$, as well.

Now by closure, $ab \in G$, but:

$ab = e \implies b = a^{-1}$ and $a = a^{-1}$, and inverses are unique, so this cannot be.
$ab = a \implies b = e$, contradiction.
$ab = b \implies a = e$, contradiction.

Hence $c = ab$.

Since $(ab)^2 = c^2 = e = ee = a^2b^2$, we have $ab = ba$, which is enough to show $G = \{e,a,b,ab\}$ is abelian.

Define $\phi: G \to \Bbb Z_2 \times \Bbb Z_2$ by:

$\phi(e) = (0,0)$
$\phi(a) = (1,0)$
$\phi(b) = (0,1)$
$\phi(ab) = (1,1)$.

It is straight-forward to verify that $\phi$ is a homomorphism by computing all 16 possible products in $G$, and evaluating $\phi$ at each one. Clearly, $\phi$ is bijective, and thus $G \cong Z_2 \times Z_2$.

If $p$ is an odd prime we have the following:

We have that $|G|=2p$, so there are $2$-Sylow and $p$-Sylow in $G$.
$$P\in \text{Syl}_p(G) , \ |P|=p \\ Q\in \text{Syl}_2(G) , \ |Q|=2$$

Let $x\in P$ and $y\in Q$, then $x^p=e$ and $y^2=e$.

Since the order of $x$ is $p$ and the order of $y$ is $2$, the subgroup that they generate has an order that divides the order of $G$, so it must be equal to the order of $G$.
Therefore, $G$ is generated by $x, y$.

If $P$ is a Sylow $p$-subgroup, and $Q$ is a Sylow 2-subgroup, we have $|PQ| = \dfrac{|P||Q|}{|P \cap Q|} = \dfrac{2p}{1} = |G|$, so we can conclude $G = PQ$, so if $x$ is a generator of $P$ and $y$ a generator of $Q$, $x$ and $y$ generate $G$. The important difference between what I wrote and what you wrote, is that my way of phrasing things precludes either of $x,y$ being the identity.

Since $[G:\langle x\rangle ]=\frac{|G|}{|\langle x\rangle |}=\frac{2p}{p}=2$ we have that $\langle x\rangle \trianglelefteq G$.

It's probably simpler just to say $P$ is normal because it is of index 2.

Since $\langle x\rangle$ is a normal subgroup of $G$ we have that $$y^{-1}xy\in \langle x\rangle, \text{ where } y\in G \\ \Rightarrow y^{-1}xy=x^t, \text{ for some } 1\leq t\leq p-1$$

We have that $$x=e^{-1}xe=(y^2)^{-1}xy^2=y^{-1}(y^{-1}xy)y=y^{-1}x^ty=y^{-1}xyy^{-1}xy\cdots y^{-1}xy=(x^t)^t=x^{t^2}$$

So $x=x^{t^2} \Rightarrow e=x^{t^2-1}$.

Therefore, $p$ divides $t^2-1=(t-1)(t+1)$, so $p$ divides $t-1$ or $t+1$.

$$p\mid t-1 \Rightarrow t-1=0 \Rightarrow t=1 \\ p\mid t+1 \Rightarrow p=t+1 \Rightarrow t=p-1$$

This is the heart of your argument, and is very nice.

When $t=1$, we have that $xy=yx$.
We have that that the order of $xy$ is $\text{lcm}(2,p) = 2p$.
Therefore, we have a cyclic abelian group of order $2p$. So, it is isomorphic to the cyclic group $\mathbb{Z}_{2p}$, right? (Wondering)

Yes, if there exists an element $a$ in a group $G$ with order $|G|$, then $G$ is generated by $a$, and is necessarily cyclic, and thus isomorphic to $\Bbb Z_{|G|}$ ($a^k \mapsto k$ gives the explicit isomorphism).

Does it always hold that when the order of $x$ is $m$ and the order of $y$ is $n$ then the order of $xy$ is $\text{lcm}(m,n)$ ? (Wondering)

Interestingly enough, no. In fact, it isn't even true if $G$ is abelian (suppose $y = x^{-1}$?). It is true when the orders are co-prime, *and* $G$ is abelian.

When $t=p−1$, we have that $x^p=e=y^2$ and $y^{−1}xy=x^{p−1}\Rightarrow yxy=x^{−1}$. So, $x$ and $y$ generate the dihedral group $D_p$, right? (Wondering)
Is the case when $p$ is an odd prime correct? (Wondering)

Yep. It's a bit wordy, but it gets the job done.

 

[mathmari]

Interestingly enough, no. In fact, it isn't even true if $G$ is abelian (suppose $y = x^{-1}$?). It is true when the orders are co-prime, *and* $G$ is abelian.

How can we prove that it holds when the orders are co-prime and $G$ is abelian? Maybe as follows? (Wondering)

Suppose that the order of $x$ is $m$ and the order of $y$ is $n$, with $\gcd (m,n)=1$.
The order of $xy$ is the smallest power, say $i$, such that $(xy)^i=1$.
Since $G$ is abelian we have that $x^iy^i=1$.
This $i$ must be the smallest integer that is a multiple of both $m$ and $n$, so it must be $\text{lcm} (m,n)$.

 

[Deveno]

How can we prove that it holds when the orders are co-prime and $G$ is abelian? Maybe as follows? (Wondering)

Suppose that the order of $x$ is $m$ and the order of $y$ is $n$, with $\gcd (m,n)=1$.
The order of $xy$ is the smallest power, say $i$, such that $(xy)^i=1$.
Since $G$ is abelian we have that $x^iy^i=1$.
This $i$ must be the smallest integer that is a multiple of both $m$ and $n$, so it must be $\text{lcm} (m,n)$.

Yep.

 

[mathmari]

For groups of order $4$, the proposition is still true, as there are only two isomorphism classes:

$\Bbb Z_4$ and $\Bbb Z_2 \times \Bbb Z_2$ (the latter is sometimes considered a dihedral group).

Why is $\Bbb Z_2 \times \Bbb Z_2$ considered a dihedral group? (Wondering)

 

[Deveno]

Why is $\Bbb Z_2 \times \Bbb Z_2$ considered a dihedral group? (Wondering)

It's (isomorphic to) the symmetry group of a "di-gon" (a line segment).

 

[mathmari]

It's (isomorphic to) the symmetry group of a "di-gon" (a line segment).

Why does this hold? (Wondering)
I got stuck right now...

 

[Deveno]

Why does this hold? (Wondering)
I got stuck right now...

It's the group $\langle r,s : r^2 = s^2 = e; rs = sr\rangle$ where $s$ is a reflection, and $r$ is a rotation of 180 degrees.

Here, we also have $srs = r^{-1}$, but $r^{-1} = r$, so $srs = r$, that is: $sr = srss = rs$.

In this "degenerate" dihedral group, (and no other) we have that our generating reflection (which we could take to be about the $x$-axis, or the $y$-axis) commutes with our generating rotation.

We can also represent this as a group of 2x2 matrices:

$I = \begin{bmatrix}1&0\\0&1\end{bmatrix}$

$R_{180} = \begin{bmatrix}-1&0\\0&-1\end{bmatrix}$

$S = \begin{bmatrix}1&0\\0&-1\end{bmatrix}$

$R_{180}S = \begin{bmatrix}-1&0\\0&1\end{bmatrix}$

(sometimes $S$ is called $H$ for "horizontal flip" and $R_{180}S$ is called $V$ for "vertical flip").