Please please please answer asap

Mathematics

2 answers:

Liula [17]2 years ago

8 0

Answer:

the answer is c

Step-by-step explanation:

Svetlanka [38]2 years ago

3 0

Answer:

I dont wanna be wrong But if it was me i would go with B...BUUUUUT C could also be right

Step-by-step explanation:

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Plz help!!! Anyone! Plz help me with this question.

Free_Kalibri [48]

Negative means less.

Positive means more.

as a negative # moves farther away from 0, it becomes smaller.

i.e.

-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7

-5 less than 7.

7 0

3 years ago

Read 2 more answers

HELP PLS!! ASAP I need it

myrzilka [38]

No idea so sorry. :(

3 0

2 years ago

Prove that H c G is a normal subgroup if and only if every left coset is a right coset, i.e., aH = Ha for all a e G

Kaylis [27]

$\Rightarrow$

Suppose first that $H\subset G$ is a normal subgroup. Then by definition we must have for all $a\in H$ , $xax^{-1} \in H$ for every $x\in G$ . Let $a\in G$ and choose $(ab)\in aH$ ( $b\in H$ ). By hypothesis we have $aba^{-1} =abbb^{-1}a^{-1}=(ab)b(ab)^{-1} \in H$ , i.e. $aba^{-1}=c$ for some $c\in H$ , thus $ab=ca \in Ha$ . So we have $aH\subset Ha$ . You can prove $Ha\subset aH$ in the same way.

$\Leftarrow$

Suppose $aH=Ha$ for all $a\in G$ . Let $h\in H$ , we have to prove $aha^{-1} \in H$ for every $a\in G$ . So, let $a\in G$ . We have that $ha^{-1} =a^{-1}h'$ for some $h'\in H$ (by the hypothesis). hence we have $aha^{-1}=h' \in H$ . Because $a$ was chosen arbitrarily we have the desired .