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 ...
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 ...
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}\...
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 ...
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 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 ...
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 ...
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 $(x^n-1)$. ...
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 ...
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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