5
votes
Accepted
Order of groups and dimensions of representations
You do not need the notion of induced representations to prove this statement. The biggest idea needed is the orthogonality of characters. Here is my proof:
Let $\Gamma$ be an irreducible $\mathbb{C}$...
- 815
4
votes
Accepted
For a finite abelian $G$, $f: G\to G$ defined by $f(g)=g^2$ is an isomorphism iff $|G|$ is odd
Suppose $|G|$ is odd. Let $g\in{ker(f)}$, meaning that $g^2=1$. Thus $g=1$ by Lagrange, otherwise $g$ is an element of order 2 in a group of odd order.
Now, suppose $|G|$ is even. Then, $G$ admits an ...
- 398
3
votes
Accepted
Preimage of an Abelian/Cyclic tower is an Abelian/Cyclic tower
Your proof for the cyclic case is not quite right.
But first I will try to guess what Lang meant.
Here is a very simple lemma.
Lemma. Let $A$, $B$ be two groups and $g: A\to B$ be an injective ...
- 7,794
3
votes
Is my proof that every group of order 4 is abelian correct?
$\mathbb{Z}_4$ does have a homomorphism going to $\mathbb{Z}_2 \times \mathbb{Z}_2$, but that homomorphism is not surjective.
Consider $g$, a generator of $\mathbb{Z}_4$. It has order $4$, but every ...
- 8,788
2
votes
Accepted
Exact sequence of direct sums of $\mathbb{Q}/\mathbb{Z}$
Let $I$ be the image of the map $(\mathbb{Q}/\mathbb{Z})^{m + n} \to (\mathbb{Q}/\mathbb{Z})^n$. Since $\mathbb{Q}/\mathbb{Z}$ is injective, the short exact sequence $$0 \to (\mathbb{Q}/\mathbb{Z})^m ...
- 309k
2
votes
Suppose $G$ is an Abelian group of order $25$ and every element of $G$ satisfies the equation $x^{25}=e$. Prove or disprove that $G$ is cyclic
Well, $x^{|G|}=e$ holds for every element of every finite group, so your question amounts to asking whether is there any noncyclic abelian group of order $25$. And the other answers have already shown ...
- 1,295
2
votes
Accepted
Given that $(G,\!\cdot\!)$ is a group and $a \in G$ such that $axa = x^{-1}$ $ \forall x \in G$ prove that G is an Abelian group
If $x=e$, then $a^2=aea=e^{-1}=e$.
On the other hand, if $x,y\in G$, then$$ax^{-1}y^{-1}a=\left(x^{-1}y^{-1}\right)^{-1}=yx.$$But, on the other hand,$$ax^{-1}y^{-1}a=ax^{-1}aay^{-1}a=xy.$$So, you ...
- 412k
2
votes
Showing ${\rm Aut}(\Bbb Z_{70})$ is abelian
The automorphism group turns out to be $\Bbb Z_4\times \Bbb Z_6$, which is not cyclic, so your approach will not work, but it is clearly abelian. Calculating this group is, I think, most easily done ...
- 192k
2
votes
How to prove that $(G, \cdot)$ structure is an Abelian group where $G$ is a set of matrices determined by a certain rule?
The limits $|x| < 1$ and the presence of $\sqrt{1-x^2}$ suggest a trig substitution of $x = \sin\theta$ or $x = \tanh\eta$. The latter case shows more promise, giving
$$
A(\eta) = \begin{bmatrix}\...
- 20k
2
votes
For a finite abelian $G$, $f: G\to G$ defined by $f(g)=g^2$ is an isomorphism iff $|G|$ is odd
Your analysis is good. In summary, if $G$ is a (multiplicative) group, then
the map $f\colon G\to G$, $f(x)=x^2$ is
a homomorphism if and only if $G$ is abelian;
surjective if and only if it is ...
- 233k
2
votes
Accepted
Is my proof that every group of order 4 is abelian correct?
Actually, the solution given on the post is wrong solely because $f$ is actually not a homomorphism as pointed out by Mariano Suárez-Álvarez in the comment section. That's because when I try to prove $...
- 1,444
2
votes
Proving a property of groups assuming they are abelian or they have a prime divisor
If $G$ is abelian, the map $x \mapsto x^{2}$ is a homomorphism, so its image $A$ is a subgroup. The kernel $K = \{ x \in G : x^{2} = 1 \}$ is a subgroup, which by Cauchy's lemma applied twice has ...
- 67k
1
vote
Proving a property of groups assuming they are abelian or they have a prime divisor
If $G$ is abelian, then the set $A$ is a subgroup. It can't be of even order. Otherwise, a larger power of $2$ would have to divide $|G|$.
Now, why is it $n$? Because there is a unique order $2$ ...
- 1,978
1
vote
Is my proof that every group of order 4 is abelian correct?
Without any appeal to orders of elements, Cauchy's Theorem of Lagrange's or type of group: assume that $G$ has $4$ elements and is not abelian. Then we can find two non-identity elements $a,b$ that do ...
- 45.3k
1
vote
Is my proof that every group of order 4 is abelian correct?
It's not true that $G\cong \Bbb Z_2×\Bbb Z_2$. $\Bbb Z_4$ has order $4$. One group is cyclic, the other isn't. So they're not isomorphic.
If you look into a Cayley table, you will find that ...
- 7,248
1
vote
Accepted
Why is the functor of abelian groups $G\mapsto G/2G$ right exact?
If $g(b-2b')=0$ then
$b-2b'\in\operatorname{im}(f)$ hence
$\bar{b}\in \text{im}(\bar{f}).$
- 14.2k
1
vote
If $G$ is cyclic of order n, $\mathbb{Q}[G]\cong \oplus_{d|n}\mathbb{Q[\xi_d]}$
Let $G=C_n$ be a cyclic group, $g$ a generator. Let $f:\mathbb{Q}[x]\to \mathbb{Q}[G]$, $\sum a_kx^k\mapsto \sum a_kg^k$. This is a surjective ring homomorphism, and its kernel is the ideal $(x^n-1)$. ...
- 3,276
1
vote
Suppose $G$ is an Abelian group of order $25$ and every element of $G$ satisfies the equation $x^{25}=e$. Prove or disprove that $G$ is cyclic
This in your OP is not a proof, as it does not show existence of an Abelian group $G$ with $|G|=25$ and $G$ not cyclic. [In fact, every group with $25$ elements has precisely $24$ non-identity ...
- 18.4k
1
vote
Accepted
One type of integer divisor homology - what does this one measure?
I have a proof that this complex we defined is exact using simplicial homology. The set $\mathbb N$ forms an abstract simplicial complex in the following fashion. Faces are integers, and the dimension ...
- 40.1k
1
vote
Define the dual group $\hat G$ as the set of all homs. from $G$ to $\Bbb C^*$, with pointwise multiplication. Show $\hat G$ is an abelian group.
Fix $G$. We check the group axioms and then commutativity.
Closure:
Let $\varphi, \psi\in \hat G$. Let $g,h\in G$. Then
$$\begin{align}\varphi\cdot \psi:G&\to \Bbb C\setminus\{0\},\\
k&\mapsto ...
- 41.4k
1
vote
Prove that that $U(n)$ is an abelian group.
Let $ a \in Un$ then we have to show that there exists $b \in Un$ such that $a.b$ mod $n = 1$.
Let us suppose $o(a)=p \implies a^p = e $
Now if $b$ is inverse of $a$ then $a.b$ mod $n = 1$ holds i....
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