Does subgroup have element that's conjugate don't have?

Question

I'm trying to prove that

there is Group G and subgroup of G, which name is H.

$a \in G $

if H is not normal subgroup of G,

$$H - \bigcup_{H \neq aHa^{-1}} aHa^{-1}$$ has at least one element.

but I can't prove this is true or not

how can I prove it?

for example

G = S_4

H = <(1 2),(3 4)>

then H = 1234 -> 1234, 2134, 1243, 2143

if a = (1 2)

$ aHa^{-1} = 1234, 2134, 1243, 2143 = H $

if a = (2 3)

$ aHa^{-1} = 1234, 2314, 1423, 2413 \neq H $

because 1234 is on H but 2314, 1423 and 2413 is not on H

Following the discussion here, I have reopened this question (as its not obviously a duplicate), but voted to close it as the OP needs to show more effort (see above comment!). — user1729, May 01 '20 at 15:27
https://math.stackexchange.com/questions/121526/union-of-the-conjugates-of-a-proper-subgroup — kkugosu, May 01 '20 at 15:38
my question is not about group G is same as union of conjugate H — kkugosu, May 01 '20 at 15:39
@kkugosu Don't worry, we know that this question is not a duplicate, which is why it has been reopened. The issue now is that you need to show some effort: What have you tried? Where did you get stuck? — user1729, May 01 '20 at 15:45
Since this question has been re-opened, I closed the duplicate question pointing to here (this one has more detail in the question). — Arturo Magidin, May 01 '20 at 16:02

score 2 · Answer 1 · answered May 01 '20 at 16:54

2

This is not true in general. It is false for the subgroup $\langle (1,2)(3,4),(3,4)(5,6) \rangle$ of $S_8$ (which has order 4), for example, but there may be smaller examples.

answered May 01 '20 at 16:54

Derek Holt

90,008

score 0 · Answer 2 · answered May 01 '20 at 19:06

Your statement is true if $H$ is a Frobenius complement of $G$. This means that $H \cap H^g=1$ for all $g \in G-H$. To this end, observe that $H \subseteq \bigcup_{g \notin N_G(H)}H^g$ is equivalent to $H \subseteq \bigcup_{g \notin N_G(H)}(H \cap H^g)$.

Does subgroup have element that's conjugate don't have?

2 Answers2