Let $G$ be a finite group, $N$ be a normal subgroup. Suppose $|G|/|N| = 100$. Prove that for every $g$ in $G$, $g^{100}$ is in $N$.

Question

Let $G$ be a finite group, $N$ be a normal subgroup. Suppose $|G|/|N| = 100$. Prove that for every $g$ in $G$, $g^{100}$ is in $N$.

How can i prove this question?

Thank you

score 2 · Answer 1 · answered Dec 02 '13 at 03:48

2

Hint: Given $g \in G$, the element $g + N \in G/N$ satisfies

$$(g + N)^{100} = e_{G/N} = N$$

Alternatively,

$$g^{100} + N = N$$

answered Dec 02 '13 at 03:48

This is based on Lagrange's Theorem and one of its corollaries isn't it? – Tristen Dec 02 '13 at 03:49
@Tristen Yes, it is. – Dec 02 '13 at 03:50

score 0 · Answer 2 · answered Dec 02 '13 at 03:47

0

Hint Consider the factor group $\frac{G}{N}$. How many elements does it have?

What can you say in $\frac{G}{N}$ about $(g+N)^{100}$?

answered Dec 02 '13 at 03:47

N. S.

2 Answers2