I want to prove or disprove this statement: If $H = Z(G),$ then $Z(G/H) = 1.$ I have seen an example here  that shows that in some cases that is correct. I want to know if this statement is correct in general or not. and if it is correct how can I prove it? Could anyone help me with that, please?
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Question. Prove or disprove: The center of $G/Z(G)$ is trivial for all finite groups $G$. (Here $Z(G)$ denotes the center of $G$.)  Attempt at a Proof.  Assume the contrary that $G/Z(G)$ has a non-trivial center.   Let $g\notin Z(G)$ be such that $gZ(G)$ is in the center of $G/Z(G)$.   Then $$\boxed{xgx^{-1}\in gZ(G)}$$ for all $x\in G$.   We note that $xgx^{-1}=ygy^{-1}$ if and only if $xC_G(g)=yC_G(g)$, where $C_G(g)$ denotes the centralizer of $g$ in $G$.  Since $g\notin Z(G)$, we have $G\neq C_G(g)$.  I couldn't go any further.  Can someone help?
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My challenge is      Prove that if $n = 0,1,2,\ldots$ and $n\Bbb Z = \lbrace nk: k \in \Bbb Z \rbrace$, show that $n\Bbb Z$ is a subgroup of $\Bbb Z$ and are the only subgroups.   I handled the first point by saying that  i) Identity is 0, and $0\in n\Bbb Z$ and $0 \in \Bbb Z$  ii) For any $a,b \in n\Bbb Z$ $a+b = na_2 + nb_2 = n(a_2+b_2) \in n \Bbb Z$  iii) For any $na \in n\Bbb Z$, its inverse $n(-a) \in n\Bbb Z$.  For the second point, I said that all of the subgroups should be infinite, because for any finite group the largest element $a+a \notin H$, where $H$ is the potential subgroup, which is a contradiction. So $H$ must be of infinite order. For any $a \in H, -a$ should also be in $H$. The difference between any $a,b \in H$ should also be in $H$, i.e. $(a-b) \in H$. This leaves only the whole group or any group of the form $n\Bbb Z$.  Is my proof complete?
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Please excuse the selfishness of the following question:     Let $G$ be a group and $H \le G$ such that $|G:H|=2$. Show that $H$ is normal.   Proof:  Because $|G:H|=2$, $G = H \cup aH$ for some $a \in G \setminus H$.   Let $x\in G$. Then $x \in H$ or $x \in aH$.  Suppose $x \in H$. Then $xhx^{-1} \in H$ because $H$ is a group.  Suppose $x \in aH$. Then $ahH(ah)^{-1} = ahHh^{-1}a^{-1} = a(hHh^{-1})a^{-1}$ ***1  ***1: Can I say now that $x \in H$, based on the that $hHh^{-1} \in H$ and that $aa^{-1} = e$ ?  Thank you for the time you spent reading my doubts.
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Let $g_n= {2^2}^n +1 $. Prove $g_0 · g_1 · · · g_{n−1} = g_{n} − 2$.    I'm not sure how to start this proof, if it should be done algebraically or if I should try to use a proof by induction.
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If $G$ is cyclic, then $G/H$ is cyclic?   The proof I got goes like this: $G$ is cyclic, so $G=&lt;g&gt;$ for some $g\in G$. So any coset in $G/H$ would be of the form $Hg'=Hg^n$ for some $n$. So $Hg$ is an generator of $G/H$. Thus, $G/H$ is cyclic.  I might just be confusing myself, but we have only shown that $Hg'$ is in form of $(Hg)^n$. But what if we are missing some $n\in\mathbb{N}$? That is, there is no quotient group of the form $Hg^2$, for example.  To make myself a little bit clearer, I think what the above proof has done was showing that $\forall Hg'\in G/H, Hg'\in &lt;Hg&gt;$, thus $G/H \subset &lt;Hg&gt;$. I feel that this is not a complete proof.
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Let $H$ be a subgroup of $G$. Prove that $gHg^{-1}=\{ghg^{-1}|h \in H\}$ is a subgroup of $G$, $g\in G$. Show that $o(gHg^{-1})=o(H)$.   Attempt:  As $e=geg^{-1}\in gHg^{-1}$ then $gHg^{-1}$ is non-empty. Let $x=gh_1g^{-1}$, $y=gh_2g^{-1}\in gHg^{-1}$, $h_1, h_2\in H$   then $xy^{-1}=\cdots=g(h_1h_2^{-1})g^{-1}\in gHg^{-1}$ as $h_1h_2^{-1}\in H$ ($H$ being the subgroup of $G$)  Therefore, $gHg^{-1}$ is a subgroup.   The problem is in 2nd part. I am unable to sgow that $o(gHg^{-1})=o(H)$. Please help me to solve the problem using the elementary tools of abstract algebra.
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I am having a hard time finding a counterexample for the statement: $G \times K \cong H \times K \implies G \cong H$  I think this should be true for abelian, finite groups. But is this true in general? What would be a counterexample?  Any hints appreciated!  Thanks
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Given a finite group $G$ and its subgroups $H,K$ such that $$G \times H \cong G \times K$$ does it imply that $H=K$.  Clearly, one can see that this doesn't work out for all subgroups. Is there any condition by which this can remain true.
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I was reading a text book and came across the following:  Important Results (This comes immediately after LCM:) If 2 [integers] $a$ and $b$ are given, and their $LCM$ and $HCF$ are $L$ and $H$ respectively, then $L \times H = a \times b$  Can some please help me understand why the above result is true? Thanks in advance.
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Let $H$ be a subgroup of a group $G$. The I know that for $a,b\in G$ we have $aH = bH$ if and only if $a^{-1}b \in H$.  My question is if it is also true that $aH = bH$ if and only if $ab^{-1} \in H$? Does it matter where the $-1$ goes?  I am guessing that this is not true since we keep putting the $-1$ on the left element, but I am also thinking that it might be true because they look like the same.  If $G$ is abelian, then it doesn't matter, so a possible counter example would have to involve a non-abelian group. If $H$ is a normal subgroup, then it should also be true because then $a^{-1}bH = Ha^{-1}b$.
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I know that given a group $G$ and two normal subgroups $H,K \subset G$ then it is not true that:  "if $H \cong K$ then $ \frac{G}{H} \cong \frac{G}{K} $ (the counterexample is quite easy with products of cyclic groups) "  My question is: Is the converse true?   i.e.     Given that $\frac{G}{H} \cong \frac{G}{K}$ then $H \cong K$ ?   I feel that the answer is no, but I can't think of an example.
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Let n be a positive integer and let G be an Abelian group. Define  $H := \{x ∈ G : |x| \ \text{divides} \ n\}$  where x denotes the order of |x|.  Show that H is a subgroup of G.  I know I need to prove that H is non-empty, contains the identity, e, and that it contains $ab^{-1}$.  I'm just not sure how to prove this, thank you for any help!
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Prove if $|z| &lt; 1$ and  $ |w| &lt; 1$, then $|1-zw^*| \neq 0$ and $| {{z-w} \over {1-zw^*}}| &lt; 1$Given that $|1-zw^*|^2 - |z-w|^2 = (1-|z|^2)(1-|w|^2)$I think the first part can be proven by saying $|1-zw^*| = 0$ if and only if $zw^*$ = 1.  And given the conditions that cannot be true. However I don't know if this part is right.
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Let $G$ be a group, and let $H$ be a subgroup of $G$. Let $x,y\in G$ be such that $xH\subseteq Hy$. Can we conclude that $xH=Hy$? this certainly holds if $H$ is finite because both sets will have the same finite cardinality. But what about the case $|H|=\infty$? I am especially interested in the case $x=y$.  EDIT: As Thomas Andrews pointed out in the comments, it is suffices to consider the case $x=y$, because the hypothesis $xH\subseteq Hy$ implies $x\in Hy$, hence $Hy=Hx$. And so, the question becomes:     Can a subgroup (of a group, of course) strictly contain some of its conjugates?
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Prove or disprove 'If $\gcd(a,b)=1$ then, $\gcd(a^2,b^2)=1$, with $a,b\not= 0$'  I need to prove this statement. I think it is true and also the converse is true.  I took some examples such as $\gcd(2,5)=1$, and $\gcd(4,25)=1$...  But how can I prove this? I know that $1=ax+by$ for some $4x,y\in \mathbb{Z}$.. But not sure   How to go from there..   Any clue?
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So i have to prove if $f(z_1)=f(z_2)$ then $z_1=z_2$ .  I tried substituting $z=x+yi$ i also tried proving their differences in absolute value are zero.None of this work .Im sure im missing a property or something that will make this very easy. I also tried making the first part of the eq belonging is $R$  so the  second part of the eq must have imaginary part  zero. I just dont get to the result i want.
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I was given a group and I have to prove   a) If $f\cdot g = f$ or $g\cdot f = f$ then $g = 1$.  Is it right to do it using just Identity axiom of group:  $f \cdot g = f = f \cdot1 \Longrightarrow g = 1 ?$  Thanks.