I encountered the following math problem, so I typed in on Mathematics Stack Exchange. Then, a list of similar posts appeared.
Let $E = {1,\dots,p}$, where $p$ is a prime number. $G$ is a transitive subgroup of $S_p$, and $H$ is a nontrivial normal subgroup of $G$. Show that $H$ acts transitively on $E$.
I notice that $\lvert G \cdot x \rvert = \lvert E \rvert = p$ because $G$ acts transitively on $E$. Now, I try to show the same for $H$, but I am stuck at $\lvert H \cdot x \rvert$. May I say that for all $g \in G$, $\lvert gH \cdot x \rvert = \lvert H \cdot x \rvert$? Why can’t $H$ fix $x$?
Is it a possible duplicate of another question?
If the answer is yes, then my post is going to be quickly flagged as duplicate by experienced users on this site, especially those who are strong at abstract algebra, and I’ll lose reputation for that.
I’ve really found the same problem on Mathematics Stack Exchange by luck.
I viewed this question four days ago. When I tried to access it again tonight from my web browsing history, I found this a bit hard. Therefore, as a post on Mathematics Meta stack Exchange suggests, Vote early, vote often.
]]>I saw a question about determinants.
Suppose that $A$ and $B$ are singular and nonsingular matrices respectively. Simplify $\det((A+B)^2−(A−B)^2)$.
A wrong solution with a vote of -2 is chosen by Daniel. Why can this happen?
That’s because he’s correctly done the expansion until $\det(2AB + 2BA)$.
Having spent time on typing a comment, I worry that it will automatically disappear in sooner or later if the accepted answer is deleted. Therefore, I back it up here.
Consider However, if $A = 0$ and $B = I_3$, then the answer is clearly zero. As a result, we can’t decude further from $\det(2(AB + BA))$.
The generation of a random matrix/array of integers using
randi([imin, imax], m, n)
. For more details, you may read
GNU Octave’s manual.
Two years ago, I thought about a group of 689 elements.^{1} I only managed to show the existence of such a group.
Inspired by the use of Sylow III to show that a group of order 15 has only one structure: $\Z_{15} \cong \Z_3 \times \Z_5$, I wondered if $\Z_{689}$ is the only possible structure for a group of order 689.
Denote $G$ as a group of order 689. First, apply Sylow III to $\abs{G}$, whose largest prime divisor 53. Since $\lvert G \rvert$ is a product of two primes 13 and 53, we conclude that $n_{53} = 1$. Second, we have
Since $53 \equiv 1 \pmod{13}$, we have to consider two cases:
is reduced to $e_G$, implying that $H$ commute with $K$. Therefore, an internal direct product of $H$ and $K$ can be set up, and $\abs{H} \times \abs{K} = 13 \times 53 = \abs{G}$, so $G = H \times K \cong \Z_{13} \times \Z_{53}$.
$n_{13} = 53$: In this case, $G$ has 53 distinct 13-Sylows $H_1, H_2, \cdots, H_{53}$. Each of these $H_i$’s is isomorphic to the cyclic group $C_{13}$. Thus, $G$ possess $53 \times 12$ elements of order 13, 52 elements of order 53, and the identity—they add up to $\abs{G}$.
Not knowing how to continue, I finally searched for “non-abelian group of order $pq$” and found this answer on Mathematics Stack Exchange, which showed that my guess is wrong. In fact, $\Aut H \cong \Z_q^*$ and its elements are $\gamma_\lambda: h \mapsto \lambda h$. $\phi(K) = \psi(K)$ because they are both subgroup of order $p$, but it’s unique in $\Aut H$. Since homomorphisms $\phi,\psi: K \to \Aut H$ are supposed to be non-trivial, $\ker \phi, \ker \psi \ne \abs{K} = p$, so they are injective. This proves the existence of a non-abelian group $G \cong H \rtimes_\psi K$.
To conclude, even though the arguments are “too simple” for a mathematicien, a group of order 689 is never simple.
In my notes, the external semidirect product $G_1 \rtimes_\gamma G_2$ of two groups $G_1$ and $G_2$ with respect to a homomorphism $\gamma: G_2 \to \Aut G_1$, is defined as
\begin{multline} \forall\, x_1,y_1 \in G_1, \forall\, x_2,y_2 \in G_2, (x_1,x_2) \times_{G_1 \rtimes_\gamma G_2} (y_1,y_2) \\ = (x_1 \times_{G_1} \gamma(x_2)(y_1), x_2 \times_{G_2} y_2). \end{multline}
Why don’t we write $(x_1,y_1)$ and $(x_2,y_2)$ instead?
I don’t think that this question can last for a day on Mathematics Stack Exchange.
If we really do so, we’ll create a chaos of indices: we've $y_1 \in G_2$ and $x_2 \in G_1$, so we substitute $y_{\color{red}{1}}$ in $\gamma: G_{\color{red}{2}} \to \Aut G_1$ and then $x_{\color{blue}{2}}$ in $\gamma(y_1): G_{\color{blue}{1}} \to G_{\color{blue}{1}}$, which belongs to $\Aut G_{\color{blue}{1}}$.
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