# Tag Info

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
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
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

### 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
Accepted

• 20k
### 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 ...