In general, and hence you first to last step doesn't work.

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- Mar 29th 2014, 05:59 AM #1

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## Plz help me on this.....

Q1:- Prove that a subgroup H of G is normal subgroup of G,

if and only if

product of any two right cosets of H in G is again a right coset of H in G.

dnt know wether i can write this or not....???

Q2:- If H is a non-empty subset of G such that Hx = HxH for every xЄG,

Then Hx = xH.

Q3:- If H and K are two normal subgroups of a finite group G such that O(H) and O(K) are

co-prime.

Then show that hk = kh for every hЄH, kЄK .

- Mar 29th 2014, 10:39 AM #2

- Mar 30th 2014, 03:02 PM #3

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## Re: Plz help me on this.....

Hi,

Some preliminary observations:

Let $G$ be any group and $A, B$ subsets of $G$. The product set is defined by $AB=\{ab:a\in A \text{ and }b\in B\}$. If A or B is a singleton, write $aB$ or $Ab$ for $AB$. Let $C$ be another subset, easily: if $A\subset B$, then $AC\subset BC$ (also you can multiply by C on the left). So if $A=B$, $AC=BC$.

1. Let $H$ be a subgroup of $G$. Then $H$ is normal in $G$ if and only if $xH=Hx$ for all $x\in G$.

a) Assume $H$ is normal in $G$ and $x\in G$. Then $x^{-1}Hx=H$. Thus $xx^{-1}Hx=xH$ and so $Hx=xH$.

b) Assume $Hx=xH$ for all $x\in G$. Let $x\in G$. Then $x^{-1}Hx=x^{-1}xH=H$. So $H$ is normal in $G$.

2. Assume $H$ is a non-empty subset of G with $Hx=HxH$ for all $x\in G$. Then $xH=Hx$ for all $x\in G$.

I can't work this problem (I may be overlooking something that is obvious). Some comments, though ( 1 is the identity of $G$):

Suppose $1\in H$. Let $x\in G$. Then $1xH\subset HxH=Hx$ and so $xH\subset Hx$. So $H=x^{-1}xH\subset x^{-1}Hx$ for any x. Choosing $x^{-1}$, I get $H\subset (x^{-1})^{-1}Hx^{-1}$ and so $H\subset xHx^{-1}$ and then $x^{-1}Hx\subset H$. Thus $H=x^{-1}Hx$. By the same argument as in 1a), I get $Hx=xH$.

Next $H1=H1H$ and so $H=HH$; i.e $H$ is closed under multiplication. So if $G$ is finite or even if $H$ contains an element of finite order, then $1\in H$ and I'm done. So a counter example must be an infinite non-abelian group with no elements of finite order in $H$. I haven't been able to come up with such an example.

3. Hints: Determine the subgroup $H\cap K$. Next for $h\in H, k\in K$, examine the element $h^{-1}k^{-1}hk$.

- Mar 31st 2014, 07:40 AM #4

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- Mar 31st 2014, 08:37 AM #5

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- Mar 31st 2014, 08:40 AM #6

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## Re: Plz help me on this.....

No, that's not right. Suppose Ha = Hb. This means that:

ha = h'b, for some h,h' in H. Yes, we CAN multiply both sides by b^{-1}, to get:

h(ab^{-1}) = h', but....

from this, all we can conclude is:

ab^{-1}in h^{-1}H, and there is NO guarantee that this equals H (if H is NOT a subgroup, then it probably ISN'T).

I think there must be a typo in the problem, because subsets H have different properties than subgroups.