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Let $ G $ be a group and $ H $ subgroup of $ G $. Is $ H $ normal? $$ 1) \ G = (\mathbb{R}^4, +), \ H = \{(x,y,z,t) \in \mathbb{R}^4 : x + 2y + 3z + 4t = 0 \}$$ $$ 2) \ G = S_4, \ H \ generated \ by \ cycle \ (1,2,3)$$ $$ 3) \ G = S_4, \ H \ generated \ by \ cycle \ (1,2,3,4)$$ $$ 4) \ G = S_5 \times S_5, \ H = S_5 \times {id} $$ $$ 5) \ G = S_{10} \times S_{10} \times S_{10}, \ H = S_{10} \times {id} \times S_{10}$$

Exists some quick method to check this? I must check condition $ \forall g \in G \ gHg^{-1} = H $ or $ gH = Hg $ but I have problem with these examples. For first example I wrote: $(a,b,c,d)H(a,b,c,d)^{-1} = \{a \cdot x \cdot a^{-1}, b \cdot y \cdot b^{-1}, c \cdot z \cdot c^{-1}, d \cdot t \cdot d^{-1}) \in \mathbb{R}^4$ it's enough?

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closed as off-topic by T. Bongers, user1729, Amzoti, Steve D, Gerry Myerson Aug 21 '13 at 13:24

This question appears to be off-topic. The users who voted to close gave this specific reason:

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The first group is abelian hence... This looks much more like homework, show some more effort – Dominic Michaelis Aug 20 '13 at 13:28
In an abelian group all subgroups are normal. What have you tried for the rest? – LASV Aug 20 '13 at 13:30
What have you learned from the answers to your earlier, similar questions?… – Gerry Myerson Aug 20 '13 at 13:30
You also have to prove that $H$ is a subgroup (unless this is defiantly given - I cannot quite tell from the first line). For the first example, you have to tell us what $\{a⋅x⋅a^{−1},b⋅y⋅b^{−1},c⋅z⋅c^{−1},d⋅t⋅d^{−1})\}$ as actually is. It turns out that it is in $H$, but you need to think a bit more about why. (Once you get to know a bit more you will hear the word abelian, and realise that this one is pretty easy!) Also, you might find this one slightly easier if you were to write it as $\{a+x-a,b+y-b,c+z-c,d+t-d)\}$ – user1729 Aug 20 '13 at 13:31
(Also, I think you need to go and think about your questions. It is difficult to help you if there is a massive stream of comments with you asking for clarification all the time (and so I am voting to close this question). Sit down and think about, for example, Q3. Come back tomorrow only asking about that one, and with some very specific questions. And play around a bit with all of this stuff until you have a better feel for it! Study the definitions and don't worry so much about getting stuff wrong...) – user1729 Aug 20 '13 at 14:04
up vote 3 down vote accepted

I can just give you a simple way. You'd better reflect about Don's again and again. We know that:

$$S_4=\{id, (3,4), (2,3), (2,3,4), (2,4,3), (2,4), (1,2), (1,2)(3,4), (1,2,3), (1,2,3,4), (1,2,4,3), (1,2,4), (1,3,2), (1,3,4,2), (1,3), (1,3,4), (1,3)(2,4), (1,3,2,4), (1,4,3,2), (1,4,2), (1,4,3), (1,4), (1,4,2,3), (1,4)(2,3)\}$$ and $$H=\langle (1,2,3)\rangle=\{id, (1,2,3), (1,3,2)\}$$ Also, you may know that if $\sigma\in S_4$ and $(k_1,k_2,...,k_3)\in S_4$ then $$\sigma^{-1}(k_1,k_2,...,k_r)\sigma=(k_1\sigma,k_2\sigma,...,k_r\sigma)$$ Now, lets try the cycles in which for example $4$ is contained in $(k_1\sigma,k_2\sigma,...,k_r\sigma)$'s for $(k_1,k_2,...,k_r)\in H$ and some $\sigma\in S_4$. This is just a feeling to get a contradiction. Of course, we are walking on the way of showing the normality. Now, take $$\sigma=(3,4)\to \sigma^{-1}=(3,4)$$ and $$\sigma^{-1}(1,2,3)\sigma=(1,2,4)\notin H$$

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Please tell me if I was wrong in any steps. I know this is an elementary approach and so I don't insist that this way is the best way and so it should be acceptable. I am here to learn more and more points and to fresh my little knowledge about Maths. Thanks. – Babak S. Aug 20 '13 at 14:11
Could you write me how you find counter-example? – Mat Aug 20 '13 at 14:12
@Mat: Just feeling. As user1729 commented in his last one, working with symmetric groups, especially $S_3, S_4$, gives us an insight and without doing exercises on them you cannot guess where to attack the problem. – Babak S. Aug 21 '13 at 10:46
$\ddot\smile \quad +1 \quad$ – amWhy Aug 21 '13 at 16:13

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