# Tag Info

## Hot answers tagged finite-groups

4

According to Maple, there are a number of values of $n < 50000$ for which $N(n) = 22$: > with( GroupTheory ): > select( n -> NumGroups( n ) = 22, [seq]( 1 .. 50000 ) ); [6321, 9075, 9765, 18135, 18669, 19215, 27075, 31017, 31605, 35685, 40053, 45045, 46431, 47565, 49539] There is a conjecture that $N$ is surjective, but to my ...

3

Use Cauchy's Theorem, which states that for every prime $p$ dividing the order of a group $G$, there exists some $g \in G$ with order $p$.

3

You can call it Multiplication table, or Addition table if you're going by addition notion. In general you can just call it Cayley Table

3

A finite group of order $n$ will have exponent $n$ if and only if its Sylow Subgroups are cyclic for every $p|n$. Suppose $p^r$ is the highest power of $p$ which divides $n$ (and $r\ge 1)$. If the Sylow subgroups associated with $p$ are not cyclic, then there is no element with order $p^r$, and hence no element with order divisible by $p^r$ and the exponent ...

2

Third attempt. In general no. $\mathbb Z_2 \times \mathbb Z_2$ is the simplest exception. Many such examples will exist. I tried to generalize but I made mistakes. I think if $n = pq$ for prime p and q. Then this will be true as some element g will have order p and only $g^{kp} = 1$ while another h will have order q and only $h^{jq} = 1$ so the only $m$ ...

2

With Holt's presentation: (we write $h^a=a^{-1}ha$, hence $h^{ab}=(h^a)^b$). Centralizer of $x$ contains $\langle x\rangle$, but it does not contain $y$ and $z$. Can it contain $yz$? $$x^{yz}=(x^y)^z=(x^9)^z=(x^z)^9=(x^{-1}y)^9\neq x$$ since there will be a factor of $y$ in the expansion [this can be seen by explicit calculation: ...

2

Think about the first theorem of homorphism. It tells us that $G/ker(\phi)\cong Im(\phi)$ so if the group $G$ is finite so you' ll get the answer.

2

Since $P \subseteq C_G(P) \subseteq N_G(P)$ and $P$ is a Sylow subgroup it follows that $|N_G(P):C_G(P)|$ is not divisible by $p$. So it must divide $p-1$. But $p$ is the smallest prime dividing the order $|G|$ and this means that $|N_G(P):C_G(P)|=1$.

2

$q\in U_p$ is its own inverse if $q^2 = 1$, that is, if $q^2 \equiv 1$ (mod $p$). This means that $p$ divides $q^2-1 = (q+1)(q-1)$, and therefore either $p\big|(q-1)$ or $p\big|(q+1)$. If $q\in U_p\setminus\{1,p-1\}$, then $q+1\in\{3\ldots p-1\}$ and $q-1\in\{1\ldots p-3\}$. $p$ is prime, and therefore it can only divide its multiples, and none of ...

2

There exists a theorem of Group Theory that says "if a prime $p$ divides the order of a finite group, this group has an element of order $p$". Thus, if all elements of $G$ have order $p$, then its order is a power of $p$.

2

$A^p(G)/G'$ is an abelian group, so if it had order divisible by $p$ then it would have a subgroup (a $p$-complement) $N/G'$ with $(A^p(G)/G')/(N/G') \cong A^p(G)/N$ a nontrivial $p$-group. Now $G/N$ is an abelian $p$-group of order larger than $G/A^p(G)$, contradiction. The proof of 2) is similar. If it was false then there would be a larger abelian ...

2

There are several examples of order $128$ in which $C$ is not abelian, such as $\mathtt{SmallGroup}(128,71)$. In this example, the maximal centralizers are exactly the maximal subgroups, so $C = \Phi(G)$, and is not abelian. I haven't thought about whether it is true in all finite $p$-groups that $C=\Phi(G)$.

2

$$G=S_3\times\mathbb{Z_2}$$ is not abelian and not cyclic hence it's not isomorphic to $\mathbb{Z_{12}}$ and $\mathbb{Z_6}\times\mathbb{Z_2}$. Also the group $A_4$ doesn't have an element of order $6$ but $G$ has, say $\{(1,2,3),1\}$, so correct option is $D_6$

1

$D_6$ contains normal subgroups $G, H$ such that $G \cong S_3$ and $H \cong \mathbb{Z}_2$. Since $G \cap H=\{1\}$ you can conclude $D_6 \cong G \times H$. I don't know if this helps, depends on your mathematical background.

1

The fact that the action is transitive implies that $n$ divides the order of $G$. Cauchy's theorem then implies that $G$ contains an element of order $n$ because $n$ is prime. What are the elements of order $n$ in $S_n$?

1

Pick a Sylow $p$-subgroup $P$ and let $G$ act on $G/P$ by left multiplication: $p\cdot gP = pgP$. This gives a homomorphism $G\to S(G/P)=S_r$, which is clearly not identically equal to $1$. Since $G$ is simple, the kernel is thus trivial and hence $S_r$ contains an isomorphic copy of $G$ as a subgroup. So $p^m r| r!$.

1

Yes, this is a special case of confluence testing in the Knuth-Bendix completion process. It is described for polycyclic presentations (including for infinite groups) in Section 12.4 of "Handbook of Computational Group Theory" by Holt, Eick and O'Brien. For a polycyclic series of length $n$ there are about $n^3/6$ consistency conditions to check - most of ...

1

One can use the class equation for $G$ a finite group to prove the more general claim that if $p^s\mid |G|$, then $G$ contains a subgroup of order $p^s$. The class equation for $G$ reads $$|G|=|Z(G)|+\sum |G:C(x_i)|$$ for finitely many $x_1,\ldots,x_r$ representatives of the nontrivial conjugacy classes of $G$. This is, I suppose, how you've proven that ...

1

I assume you have shown that $\phi$ is a homomorphism. Showing that a homomorphism is injective is equivalent to showing that its kernel is trivial. Let $a \in \ker \phi$. We want to show that $a=1$. Let $g$ be the order of $G$ and let $k$ be the order of $a$. By Lagrange's theorem, $k \mid g$. Since $a \in \ker \phi$, we know that $a^n = 1$. Hence $k \mid ... 1 As suggested in the comments, the answer is that cycles$\sigma$and$\tau$commute if and only if either: (i)$\sigma$and$\tau$are disjoint; or (ii)$\sigma$is a power of$\tau$and$\tau$is a power of$\sigma$. In case (ii), this implies that they have the same order and hence the same length, and they must both be cycles on the same set of points. ... 1 If$\phi :G\to F$then$G/\operatorname{ker \phi}\cong \operatorname{Im\phi}=\phi(G)\implies o(G/\operatorname{ker \phi})=o(\operatorname{Im \phi})\implies \dfrac{o(G)}{o(\operatorname {Im \phi)}}=o(\operatorname{ker \phi})$and$\operatorname{ker \phi}$is a subgroup of$G$1 Using Chinese Remainder theorem you get : $$\mathbb{Z}/10\mathbb{Z}\times \mathbb{Z}/10\mathbb{Z}\text{ is isomorphic to }[\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}]\times[\mathbb{Z}/5\mathbb{Z}\times \mathbb{Z}/5\mathbb{Z}]$$ Now write : $$S_2:=\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$$$\$S_2:=\mathbb{Z}/5\mathbb{Z}\times ...

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