# Tag Info

6

Note that $$(1\,2\,3\,4\,5)(1\,2\,3\,5\,4)=(1\,3)(2\,4)$$

4

You could try Lagrange's Theorem which tells you something about the order of a subgroup. Count the elements of $H$.

4

Let $X$ be the set of all partitions of the set $\{1,2,3,4\}$ consisting of two parts with two elements each, so that $$X=\Bigl\{\big\{\{1,2\},\{3,4\}\big\},\big\{\{1,3\},\{2,4\}\big\},\big\{\{1,4\},\{2,3\}\big\}\Bigr\}.$$ If $g\in S_4$ and $\pi=\big\{\{a,b\},\{c,d\}\big\}\in X$, then $\big\{\{g(a),g(b)\},\{g(c),g(d)\}\big\}$ is another element of $X$, which ...

3

We know that $\mathbb{Z}_{19}^\times = 18$, and that $2$ generates a subgroup, $H$. By Lagrange's theorem, the order of $H$ must divide $18$, so we must have $|H| \in \{1,2,3,6,9,18\}$. We have \begin{align} 2^1 &= 2 \neq 1\\ 2^2 &= 4 \neq 1\\ 2^3 &= 8 \neq 1\\ 2^6 &= 64 \equiv 7 \neq 1\\ 2^9 &= 2^6 \cdot 2^3 \equiv 7 \cdot 8 = 56 ...

3

I agree with Mike that you should ask a faculty member about this, since we have no concrete idea about how deep your lecture program goes, nor what is of interest to your professor. That being said, there is an answer lying perfectly in the intersection of your two areas, and that is Klein Geometry. Klein started the so called Erlangen Program in the ...

3

Perhaps you've seen that the group of inner automorphisms of a group, $\text{Inn}(G)$ is a subgroup of $\text{Aut}(G)$ isomorphic to $G/Z(G)$? Now $Z(D_6) = \{e, r^3\}$, and it is not hard to show that $D_6/Z(D_6)$ is a non-abelian group of order $6$. So $\text{Aut}(D_6)$ has order at least (and a multiple of) $6$. Any automorphism must send only ...

2

Hint: To enumerate the automorphisms, recall that an automorphism is uniquely determined by how it acts on the generators of the group. Consider the presentation, $$D_{12}=\langle r,s \mid r^6=s^2=1, srs=r^{-1}\rangle = \{ 1,r,r^2,r^3,r^4,r^5,s,sr,sr^2,sr^3,sr^4,sr^5\}.$$ The subgroup $\langle r\rangle$ is the unique subgroup of index 2, which must therefore ...

2

The argument of Derek Holt shows that it suffices to show it for $p$ -groups. Now to show it for $p$-groups you need to show that the center of a non-trivial $p$-group, is non-trivial and that it holds for abelian $p$-groups. Then you show the property by induction. If $G$ is a non-abelian $p$-group then its center $Z$ is proper non-trivial. Take $p^k$ ...

2

Since $G/Z(G)$ is abelian, for any $x,y \in G$, the commutator $[x,y]=x^{-1}y^{-1}xy$ is in $Z(G)$. Now $yx=xy[y,x]$. To transform $(xy)^p$ to $x^py^p$ we have to carry out this transformation a total of $p(p-1)/2$ times, which is a multiple of $p$ because $p$ is odd.. So, using $[y,x] \in Z(P)$, we get $$(xy)^p = x^py^p[y,x]^{p(p-1)/2} = x^py^p,$$ which ...

2

For any $a,b\in G$ $$[a,b] := aba^{-1}b^{-1}$$ Since $G/Z(G)$ is abelian, one can see that $$[a,b] \in Z(G) \quad\forall a,b\in G$$ Hence, $[a,b]^p = e$ for all $a,b\in G$. Now for any $x,y\in G$, note that $$(xy)^p = x^py^p[x,y^{-1}]^{p(p-1)/2} = x^p y^p ([x,y^{-1}]^p)^{(p-1)/2} = x^py^p$$

2

The condition is equivalent to a Sylow $p$-subgroup being self-normalizing in $S_n$, and the answer is very simple (simple to state, anyway): $p=2$ is both necessary and sufficient for the condition to hold for all $n$. It is well-known that a Sylow $p$-subgroup $P$ of $S_n$ is a direct product of iterated wreath products $P_k = C_p \wr C_p \wr \cdots \wr ... 2 A way that doesn't use group actions (which are awesome, by the way, so try to use them when you can): We want to know the order of the centralizer of$g,\ |C(g)|$. Now$x \in C(g)$means that$xgx^{-1} = g$. Since$x(1\ 2\ 3)x^{-1} = (x(1)\ x(2)\ x(3))$we can immediately see two things:$x$must map$\{1,2,3\} \to \{1,2,3\}$and thus$\{4,5\} \to ...

2

Let $G$ acts on itself by conjugation. Consider the orbit of $g$, it's the conjugacy class of $g$ in $S_{5}$, so the set of $3$ cycles. Now, $H$ is the stabilizer of $g$. If $x \in H$, then $xg=gx$ so $xgx^{-1}=g$. And if $x \in Stab(g)$, then $xgx^{-1}=g$ so $xg=gx$ so $x \in H$. Now the orbit stabilizer theorem tells us that $|G|=|Orb(g)||Stab(g)|$, so ...

2

Yes. If $Z(G)\not\subseteq H$, then letting $x\in Z(G)$ will satisfy the problem. If $Z(G)\subseteq H$, then we take quotient by $Z(G)$ and apply induction. Suppose $x\mapsto \bar{x}$ denotes the quotient map. Then, by the induction hypothesis, there is $\bar{x}\in\bar{G}\backslash\bar{H}$ such that $\overline{xH}=\overline{Hx}$. It is easy to conclude ...

1

The group $A_4$ is solvable and has no subgroup of size $6$, so $A_4\times \mathbf{Z}/101\mathbf{Z}$ is a counterexample for you.

1

Since $S_4$ is generated by $u=(12)$ and $v=(1234)$, it is enough to check that $uVu^{-1}=V$ and $vVv^{-1}=V$. That solves your problem about «nor having a general formula for $x$».

1

Note: I am proving in first way, Not by Isomorphism theorem, The only proper non-trivial normal subgroup of $S_4$ are $$V_4 = \{e,(12) (34) , (13)(24), (14) (23)\}$$ and $A_4$. Suppose that $N$ is a normal proper non-trivial subgroup of $S_4$. First note that $N$ does not contain a transposition, because if one transposition $τ$ lies in $N$, then $N$ ...

1

One such example is $\langle a,b,c \mid a^{-1}ba=b^2, b^{-1}cb=c^2, c^{-1}ac=a^2 \rangle$. You can use this to construct a sequence of examples of increasing complexity. The example above is the first in the sequence and has total relator length $15$. The second group in the sequence is \langle a,b,c \mid A^{-1}BA=B^2, B^{-1}CB=C^2, C^{-1}AC=A^2 ...

1

If $A \le L$ then $[A,V] \le A \cap V = 1$. $T/A = TA/A \cong T/T \cap A$ is a $p'$-group hence so is $T/T \cap N \cong TN/N$ so $T \le F$. Since $A$ is minimal normal, $[T,A] = 1$ or $A$. If $[T,A]=1$ then $A \le Z(T)$ and $T = A \times B$ with $B$ a $p'$-group and $B \unlhd G$. So, since $A$ is the unique minimal normal subgroup, $B=1$ and $T=A$. So ...

1

If $N \ne O_p(G)$ then some prime $q \ne p$ divides $|N|$ and then $O_q(N)=O_q(G) \ne 1$ contradicting the fact that $A$ is the unique minimal normal subgroup of $G$. Note that $A \cap Z(N)$ is a nontrivial normal subgroup of $G$ contained in $A$ and, since $A$ is a minimal normal subgroup, we must have $A \cap Z(N)=A$ so $A \le Z(N)$.

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