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# Let p be a prime- Show that if H is a subgroup of a group with order 2.docx

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# Let p be a prime- Show that if H is a subgroup of a group with order 2.docx

Let p be a prime. Show that if H is a subgroup of a group with order 2p that is not normal, then H has an order of 2.

I feel that I need to use Lagrange\'s Theorem but I am not sure how to get started. Anything will help, thanks!
Solution
You are correct you need Lagrange\'s Thm. Notice the only divisors of 2p are 2 and p. This implies that the order of H is either 2,p, or 2p. If order of H = 2p then H = G and H is normal in G If order of H = p then the index of H in G (usually denoted [G:H]) is 2. This means there are only two distinct cosets of H in G. This guarantees H is normal. It is fairly straight forward to prove any subgroup of index 2 is normal. If you would like a proof write back. Thus if H is not normal it must have order 2.
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Let p be a prime. Show that if H is a subgroup of a group with order 2p that is not normal, then H has an order of 2.

I feel that I need to use Lagrange\'s Theorem but I am not sure how to get started. Anything will help, thanks!
Solution
You are correct you need Lagrange\'s Thm. Notice the only divisors of 2p are 2 and p. This implies that the order of H is either 2,p, or 2p. If order of H = 2p then H = G and H is normal in G If order of H = p then the index of H in G (usually denoted [G:H]) is 2. This means there are only two distinct cosets of H in G. This guarantees H is normal. It is fairly straight forward to prove any subgroup of index 2 is normal. If you would like a proof write back. Thus if H is not normal it must have order 2.
.

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### Let p be a prime- Show that if H is a subgroup of a group with order 2.docx

1. 1. Let p be a prime. Show that if H is a subgroup of a group with order 2p that is not normal, then H has an order of 2. I feel that I need to use Lagrange's Theorem but I am not sure how to get started. Anything will help, thanks! Solution You are correct you need Lagrange's Thm. Notice the only divisors of 2p are 2 and p. This implies that the order of H is either 2,p, or 2p. If order of H = 2p then H = G and H is normal in G If order of H = p then the index of H in G (usually denoted [G:H]) is 2. This means there are only two distinct cosets of H in G. This guarantees H is normal. It is fairly straight forward to prove any subgroup of index 2 is normal. If you would like a proof write back. Thus if H is not normal it must have order 2.