## Determine the Number of Elements of Order 3 in a Non-Cyclic Group of Order 57

## Problem 628

Let $G$ be a group of order $57$. Assume that $G$ is not a cyclic group.

Then determine the number of elements in $G$ of order $3$.

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Let $G$ be a group of order $57$. Assume that $G$ is not a cyclic group.

Then determine the number of elements in $G$ of order $3$.

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Let $G$ be a group of order $12$. Prove that $G$ has a normal subgroup of order $3$ or $4$.

Add to solve laterLet $G$ a finite group and let $H$ and $K$ be two distinct Sylow $p$-group, where $p$ is a prime number dividing the order $|G|$ of $G$.

Prove that the product $HK$ can never be a subgroup of the group $G$.

Add to solve laterLet $G$ be a finite group of order $231=3\cdot 7 \cdot 11$.

Prove that every Sylow $11$-subgroup of $G$ is contained in the center $Z(G)$.

Let $G$ be a non-abelian group of order $pq$, where $p, q$ are prime numbers satisfying $q \equiv 1 \pmod p$.

Prove that a $q$-Sylow subgroup of $G$ is normal and the number of $p$-Sylow subgroups are $q$.

Add to solve laterProve that any $p$-Sylow subgroup of a group $G$ of order $33$ is a normal subgroup of $G$.

Add to solve laterLet $p, q$ be prime numbers such that $p>q$.

If a group $G$ has order $pq$, then show the followings.

**(a)** The group $G$ has a normal Sylow $p$-subgroup.

**(b)** The group $G$ is solvable.

Suppose that $G$ is a finite group of order $p^an$, where $p$ is a prime number and $p$ does not divide $n$.

Let $N$ be a normal subgroup of $G$ such that the index $|G: N|$ is relatively prime to $p$.

Then show that $N$ contains all $p$-Sylow subgroups of $G$.

Add to solve laterLet $G$ be a finite group. Suppose that $p$ is a prime number that divides the order of $G$.

Let $N$ be a normal subgroup of $G$ and let $P$ be a $p$-Sylow subgroup of $G$.

Show that if $P$ is normal in $N$, then $P$ is a normal subgroup of $G$.

Let $G$ be a finite group and $P$ be a nontrivial Sylow subgroup of $G$.

Let $H$ be a subgroup of $G$ containing the normalizer $N_G(P)$ of $P$ in $G$.

Then show that $N_G(H)=H$.

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