Showing that $G$ is abelian, using order of $G$ and center of $G$.

Question

My question is the following: Let $G$ be a group of order $7 \times 43 \times 47$ and $Z(G)$ contains an element of order $7$. Show that $G$ is abelian.

Note: This is a question from a algebra qual exam of Maryland University January 2021.

My Attempt: I use Sylow Theorem to get the followings:

$n_{47} = 1$.

$n_{43} = 1$.

$n_7 = 1$ or $n_7 = 43$.

I try to use a case analysis:

Case1: $n_7 = 1$. In this case, all of the Sylow subgroups are normal subgroups of $G$. Thus, $G \cong\Bbb Z_7 \times \Bbb Z_{43} \times\Bbb Z_{47}$. Is this conclusion true? If yes, what is the precise statement of the theorem i am using here? I intuitively think that this is true. However, I do not have any rigorous proof or idea.

Case2: I do not know what to do there. Since I did not use the assumption $Z(G)$ contains an element of order $7$ yet. It must be used here?

Could you help me please?

Thanks in advance:)

It follows from two duplicates, namely from this one, and this one - see the answer below. — Dietrich Burde, Jun 11 '21 at 15:10
The fact that $Z(G)$ contains an element of order $7$ implies that $n_7=1$. — Derek Holt, Jun 11 '21 at 16:11
If $Z(G)$ contains an element of order $7$, then it contains a subgroup of order $7$, which is a normal Sylow $7$-subgroup of $G$. So $n_7=1$. — Derek Holt, Jun 11 '21 at 16:27

score 1 · Answer 1 · answered Jun 11 '21 at 14:18

1

Hint:

Use the following facts.

$G$ is abelian if $G/Z(G)$ is cyclic.
A group of order $pq$, where $p,q$ are primes and $p > q$, is cyclic if $q$ does not divide $p-1$.

answered Jun 11 '21 at 14:18

Ajay Kumar Nair

623

Thank you very much:) I get it know. – Ywestes Jun 11 '21 at 15:17

Showing that $G$ is abelian, using order of $G$ and center of $G$.

1 Answers1