I'm studying for finals and came across this problem in a book.  Suppose $G$ is a finite group with conjugacy class representatives $g_1,...,g_k$ and character table $Z$. Consider $Z$ as a matrix. Then I want to show that $\det(Z)$ is either real or purely imaginary and that $$|\det(Z)|^2=\prod_{i=1}^k |C_G(g_i)|,$$ where $|C_G(g_i)|$ is the centralizer of $g_i$.  I am very lost as to how to approach this problem. Any help would be greatly appreciated.
121318
This is an exercise from the book "Groups and Representations" by Alperin &amp; Bell. This quantity is well defined upto a sign. By column orthogonality relations, its squared norm is  $\displaystyle\prod_{\substack{g}} |C_G(g)|$, where g runs over the representatives of the conjugacy classes. If the group is cyclic, the determinant is just a Vandermonde determinant.   I wonder if there is a nice explanation for an arbitrary group.
25848
Question: If $G$ is a finite group such that the product of its elements (each chosen only once) is always $1$, independent of the ordering in the product, what can we say about $G$?  I was trying to conclude that $G$ must be abelian (I am not sure about this). I proceed as follows: if $|G|\leq 4$, then obviously, $G$ is abelian. Suppose $|G|&gt;4$. Then consider products of all elements in the order $a^{-1}b^{-1}ab cc^{-1}dd^{-1}....$. This, being identity, implies that $a^{-1}b^{-1}ab=1$, i.e. $G$ is abelian.  BUT, the problem is in $a^{-1}b^{-1}ab$, the elements $a$ and $a^{-1}$ should be distinct, and similarly for $b$. Otherwise, the product $a^{-1}b^{-1}ab cc^{-1}dd^{-1}....$ is not in the prescribed form as in question.  Also, note that the cyclic group of order $4$ doesn't satisfy that $abcd=1$ where $\{a,b,c,d\}\cong \mathbb{Z}_4$.
30139
Assume that $N$ is a normal subgroup of $G$ such that $N \cap G'=\{e\}$.  Prove that $N \subset Z(G)$.   Note 1: $Z(G)$ is the center of the group $G$, and $G'$ is the set of all commutators of $G$.  Note 2: Assume that $n \in N$. We want to show that for all $g \in G,$  $ng=gn$. But, I don't know how to use the fact that $N \cap G'=\{e\}$ to reach this.    Any hints? suggestions? answers?
84773
I was trying to find an example of a group $G$ and normal subgroups $N_1 , \ldots , N_n$ such that $G = N_1\cdots N_n$ and $N_i \cap N_j = \{e\}$ for all $i \neq j$ and yet $G$ is not the internal direct product of  $N_1, \ldots , N_n$ .  It is easy to show that G is the internal direct product of normal subgroups $N_1 , \ldots , N_n$ iff  (i) $G = N_1\cdots N_n$  (ii) $N_i\cap (N_1 \cdots N_{i-1}\cdots N_{i+1} \cdots N_n) = \{e\}$ for $i = 1, 2, \ldots, n$.  So I guess the question boils down to produce an example of a group $G$ and its normal subgroups  $N_1 , \ldots , N_n$ such that   (1) $G = N_1\cdots N_n$  (2) $N_i \cap N_j = \{e\}$ for all $i \neq j$   but     (3) $N_i \cap (N_1 \cdots N_{i-1} N_{i+1} \cdots N_n) \neq \{e\}$ for at least one $1 \le i \le n.$ Thanks in advance for help....
94994
Let $M$ be a (real) $n \times n$ matrix. For $1 \leq i, j \leq n$ we denote by $M_{ij}$ the $(n-1) \times (n-1)$ matrix that we get when the $i$th row and $j$th column of $M$ are removed. Now, consider fixed $i$ and $j$ with $i\neq j$. Let $N$ be the $(n-2) \times (n-2)$ matrix that we get when removing both the $i$th and $j$th row and the $i$th and $j$th column from $M$. Then the following identity holds: $$ \det M \det N = \det M_{ii}\det M_{jj} - \det M_{ij} \det M_{ji}. $$  We were able to prove this looking at all different terms that can occur on both sides when evaluating the determinant as a sum of $n!$ terms (for instance, terms containing $a_{ji}a_{ij}$ are counted both $1$ time in the LHS and $1$ time in the RHS).  We are looking for a slick proof that does not involve writing out the determinant.  Any suggestions or approaches to this problem are appreciated!
19132
I will quote a question from my textbook, to prevent misinterpretation:     Let $G$ be a finite abelian group and let $m$ be the least common multiple of the orders of its elements. Prove that $G$ contains an element of order $m$.   I figured that, if $|G|=n$, then I should interpret the part with the least common multiple as $lcm(|x_1|,\dots,|x_n|)=m$, where $x_i\in G$ for $0\leq i\leq n$, thus, for all such $x_i$, $\exists a_i\in\mathbb{N}$ such that $m=|x_i|a_i$. I guess I should use the fact that $|x_i|$ divides $|G|$, so $\exists k\in \mathbb{N}$ such that $|G|=k|x_i|$ for all $x_i\in G$. I'm not really sure how to go from here, in particular how I should use the fact that $G$ is abelian.
110351
I have seen and studied this class equation for a finite group acting on itself by conjugations. The only applications I know are Cauchys' theorm and Sylow's theorem. Are there more?
14245
Let $n&gt;2$ and $\tau\in A_n$ and let $C=\{g\tau g^{-1}|g\in S_n\}$ be the conjugacy class of $\tau$ in $S_n$.  Suppose that there is no odd element $x\in S_n$ such that $\tau x=x\tau$.  Prove that $C$ is a  union of two disjoint conjugacy classes in $A_n$.  I managed to observe that if $\sigma_1\in S_n$ an even element and $\sigma_2\in S_n$ an odd element, then:  $\sigma_1\tau\sigma_1^{-1}\neq\sigma_2\tau\sigma_2^{-1}$ otherwise we get that the odd element $x=\sigma_2^{-1}\sigma_1$ commutes with $\tau$.  So, I got one conjugacy class to be the conjugacy class of $\tau$ in $A_n$, say $Q$. It is now clear that it contains all the elements in $C$ of the form $g\tau g^{-1}$ for $g$ even. But I couldn't prove that $C\setminus Q$ can be generated by a conjugacy class in $A_n$ and moreover the sizes equivalence $|Q|=|C\setminus Q|$.
42552
Let $a$ be an element of a group G. Prove that $a$ commutes with each of its conjugates in $G$ iff $a$ belongs to an abelian normal subgroup of $G$.   My Try:  I proved the backward direction. For forward one, given $g\in G, a(gag^{-1})=(gag^{-1})a$. I was trying to prove that $a\in Z(G)$. In order to prove that, I must show that $ag=ga$. But failed. Am I on the right track? Can anybody please give me a hint? I want to try it myself.
15331
Having problems with the following statement:     If $N$ is a nontrivial normal subgroup of a nilpotent group $G$ then $N \cap Z(G) \neq \langle e \rangle$   Here $Z(G)$ denotes the center of $G$.   I've solved the finite case, where I reduced the problem to proving that a p-$group$ has this property. The infinite case resists my attempts though.   Since $G$ is nilpotent, the ascending central series ends with $G$ itself: $\langle e\rangle \subseteq Z_1(G) \subseteq \ldots \subseteq Z_n(G) = G$ where $Z_1(G) = Z(G)$. Since $N \subseteq G$, there must be a smallest $i$ such that $N \cap Z_i(G) \neq \langle e \rangle$. I would like to show that $i = 1$ in this case, or perhaps just show that $N \cap Z_i(G)$ commutes with elements of $G$. Could somebody give me a tip in the right direction?
68692
Let $V$ be a complex-vector space and $x,y\in gl(V)$ such that $[x y]=xy-yx$ commutes with $x$ and $y$. Let $[x y]$. Prove that   (1)$tr(z^{m})=0$ , for every $m\geq 1$.  (2)Deduce that z is a nilpotent map.   I couldn't do the first one. For the second one if we can prove that the possible eigen values of $z$ is only $0$, we are done. From the first one , we have if $\lambda_{1},... \lambda_{n}$ are eigen values of $z$, then $\sum_{i=1}^{n} \lambda_{i}^{k} =0$, for every $k$. From there how can we comclude that $\lambda_{1}=\lambda_{2}=...= \lambda_{n}=0$.  Thanks in advance for any help.
48825
I want to show that if $g,h\in G$ are group elements with finite coprime orders $m$ and $n$ and $gh=hg$ then the order of $gh$ is $mn$.    We have that  $$ (gh)^{mn}=g^{mn}h^{mn}=1 $$  using $gh=hg$, so $|gh|\leq mn$. On the other hand, $mn=\operatorname{gcd}(m,n)\operatorname{lcm}(m,n)=\operatorname{lcm}(m,n)$, since $m$ and $n$ are coprime. I am not sure how to proceed, I tried assuming that $|gh|&lt;mn$ and arriving at a contradiction, but it did not take me anywhere.
81509
Let $G$ be a group s.t. every non-identity element has order 2. If $G$ is finite, prove $|G| = 2^n$ and $G \simeq C_2 \times C_2 \times\cdots\times C_2$  I know G is abelian since $ab = (ab)^{-1} = b^{-1} a^{-1} = ba$ for all non-trivial $a,b \in G$ so I have several questions remaining:    How do I prove $|G| = 2^n$? I'd like to say we use induction to prove this but I'm at a loss as to where I would start. Why is $G \simeq C_2 \times C_2 \times\cdots\times C_2$?   I've read several answers to question similar to this yet unfortunately most of them involve Galois Fields and vector spaces, both concepts I'm unfamiliar with. I'd greatly appreciate an intuitive proof.
5906
There is the standard proof using $$\det(A)=\det( A^{T} ) = \det(-A)=(-1)^n \det(A)$$  I would like a proof that avoids this. Specifically, there is the proof that for $A$ a $\bf{real} $ matrix, the transpose is the same as the adjoint, which gives (using the complex inner product) $\lambda \|x\|^2 =\langle  Ax, x \rangle= \langle  x, -Ax \rangle=-\overline{\lambda } \|x\|^{2}$, so any eigenvalue is purely imaginary. Then we conclude that, since any odd-dimensional real matrix has a real eigenvalue, that eigenvalue must be zero.  This argument doesn't work for a general complex skew-symmetric matrix.  Is there something I'm missing, is there a way to modify this argument to get that zero is an eigenvalue for the complex case?  Also, can somebody please give a geometric reason why odd-dimensional skew-symmetric matrices have zero determinant (equiv., a zero eigenvalue)?  Thanks!
74690
I need help to answer the folowing problem:  Let $F$ be a field, $n\ge 1$ and $G=F^n\bigoplus F^n\bigoplus Mat(n,F).$ That is an element of $G$ has the form $(v,w,X)$ where $v$ and $w$ are n-by-1 vectors and $X$ is $n$-by$-n$ matrix, all with entires in $F$. Define an operator on $G$ by: $$(v_1,w_1,X_1)(v_2,w_2,X_2)=(v_1+v_2,w_1+w_2,X_1+X_2+v_1w_2^t)$$ Show that $G'=\left\{(0,0,X), X\in Mat(n,F)\right\}$. Where $G'$ is the derived group of $G$ i.e $G′=\left\{[a,b],a,b∈G\right\}$
26707
Let $G$ be the group of all $n\times n$ matrices with entries of the matrices from the field $\mathbb Z_p$ , $p$ prime, such that determinant of every matrix is not $[0]$ , w.r.t. matrix multiplication ; then what is the order of $G$ ?
54848
For fixed prime power $q$ show that the general linear group $GL_n(\mathbb F_q)$ of invertible matrices with entries in the finite field $\mathbb F_q$ has an element of order $q^n-1$.   I tried to show this question with showing diagonal matrix but i can't find element directly  competible with order i think i am on wrong way please give me clue ?