Union of finite subgroups not necessarily normal?

Question

Suppose G is a group and n is a natural number. Show the union of subgroups of G of order n is not necessarily a normal subgroup of G.

I'm guessing I use proof by contradiction, but I have no idea where to begin. Do I use the trivial subgroup?

Hint: The union of two subgroups $H_1,H_2 \subset G$ is a subgroup only if one is contained in the other. — hardmath, Oct 10 '16 at 22:10
Why on earth do you think a proof by contradiction is required? To show that some proposition $\phi$ is not necessarily true, what you do is produce a counter-example to $\phi$ (as those who have answered your question have done). — Rob Arthan, Oct 10 '16 at 22:11

score 3 · Accepted Answer · answered Oct 10 '16 at 21:56

3

It may we be that union is not even a subgroup. For example, the union of all the subgroups of order two in $\;S_3\;$ :

$$\left\{\;(1),\,(12),\,(13),\,(23)\;\right\}\rlap{\;\,/}< S_3$$

answered Oct 10 '16 at 21:56

DonAntonio

score 2 · Answer 2 · answered Oct 10 '16 at 21:56

2

Hint: Consider $G=S_3$ and $n=2$. - Or even more trival: Consider any finite $G$ and $n$ not a divisor of $|G|$

answered Oct 10 '16 at 21:56

Hagen von Eitzen

2 Answers2