Let $G$ be a finite group of order $2n$.
Suppose that exactly a half of $G$ consists of elements of order $2$ and the rest forms a subgroup.
Namely, suppose that $G=S\sqcup H$, where $S$ is the set of all elements of order in $G$, and $H$ is a subgroup of $G$. The cardinalities of $S$ and $H$ are both $n$.
Then prove that $H$ is an abelian normal subgroup of odd order.
Let $G$ be a finite group and let $A, B$ be subsets of $G$ satisfying
\[|A|+|B| > |G|.\]
Here $|X|$ denotes the cardinality (the number of elements) of the set $X$.
Then prove that $G=AB$, where
\[AB=\{ab \mid a\in A, b\in B\}.\]
Let $G$ be a finite group and let $S$ be a non-empty set.
Suppose that $G$ acts on $S$ freely and transitively.
Prove that $|G|=|S|$. That is, the number of elements in $G$ and $S$ are the same.
In this post, we study the Fundamental Theorem of Finitely Generated Abelian Groups, and as an application we solve the following problem.
Problem.
Let $G$ be a finite abelian group of order $n$.
If $n$ is the product of distinct prime numbers, then prove that $G$ is isomorphic to the cyclic group $Z_n=\Zmod{n}$ of order $n$.
Let $R$ be a commutative ring with $1$ and let $G$ be a finite group with identity element $e$. Let $RG$ be the group ring. Then the map $\epsilon: RG \to R$ defined by
\[\epsilon(\sum_{i=1}^na_i g_i)=\sum_{i=1}^na_i,\]
where $a_i\in R$ and $G=\{g_i\}_{i=1}^n$, is a ring homomorphism, called the augmentation map and the kernel of $\epsilon$ is called the augmentation ideal.
(a) Prove that the augmentation ideal in the group ring $RG$ is generated by $\{g-e \mid g\in G\}$.
(b) Prove that if $G=\langle g\rangle$ is a finite cyclic group generated by $g$, then the augmentation ideal is generated by $g-e$.
Use Lagrange’s Theorem in the multiplicative group $(\Zmod{p})^{\times}$ to prove Fermat’s Little Theorem: if $p$ is a prime number then $a^p \equiv a \pmod p$ for all $a \in \Z$.
Let $G$ be a finite group of order $n$ and let $m$ be an integer that is relatively prime to $n=|G|$. Show that for any $a\in G$, there exists a unique element $b\in G$ such that
\[b^m=a.\]
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